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#include<bits/stdc++.h>
using namespace std;
 
// https://v...content-available-to-author-only...e.com/cp-geo.pdf
const int N = 3e5 + 9;
 
const double inf = 1e100;
const double eps = 1e-9;
const double PI = acos((double)-1.0);
int sign(double x) { return (x > eps) - (x < -eps); }
struct PT {
    double x, y;
    PT() { x = 0, y = 0; }
    PT(double x, double y) : x(x), y(y) {}
    PT(const PT &p) : x(p.x), y(p.y)    {}
    PT operator + (const PT &a) const { return PT(x + a.x, y + a.y); }
    PT operator - (const PT &a) const { return PT(x - a.x, y - a.y); }
    PT operator * (const double a) const { return PT(x * a, y * a); }
    friend PT operator * (const double &a, const PT &b) { return PT(a * b.x, a * b.y); }
    PT operator / (const double a) const { return PT(x / a, y / a); }
    bool operator == (PT a) const { return sign(a.x - x) == 0 && sign(a.y - y) == 0; }
    bool operator != (PT a) const { return !(*this == a); }
    bool operator < (PT a) const { return sign(a.x - x) == 0 ? y < a.y : x < a.x; }
    bool operator > (PT a) const { return sign(a.x - x) == 0 ? y > a.y : x > a.x; }
    double norm() { return sqrt(x * x + y * y); }
    double norm2() { return x * x + y * y; }
    PT perp() { return PT(-y, x); }
    double arg() { return atan2(y, x); }
    PT truncate(double r) { // returns a vector with norm r and having same direction
        double k = norm();
        if (!sign(k)) return *this;
        r /= k;
        return PT(x * r, y * r);
    }
};
istream &operator >> (istream &in, PT &p) { return in >> p.x >> p.y; }
ostream &operator << (ostream &out, PT &p) { return out << "(" << p.x << "," << p.y << ")"; }
inline double dot(PT a, PT b) { return a.x * b.x + a.y * b.y; }
inline double dist2(PT a, PT b) { return dot(a - b, a - b); }
inline double dist(PT a, PT b) { return sqrt(dot(a - b, a - b)); }
inline double cross(PT a, PT b) { return a.x * b.y - a.y * b.x; }
inline double cross2(PT a, PT b, PT c) { return cross(b - a, c - a); }
inline int orientation(PT a, PT b, PT c) { return sign(cross(b - a, c - a)); }
PT perp(PT a) { return PT(-a.y, a.x); }
PT rotateccw90(PT a) { return PT(-a.y, a.x); }
PT rotatecw90(PT a) { return PT(a.y, -a.x); }
PT rotateccw(PT a, double t) { return PT(a.x * cos(t) - a.y * sin(t), a.x * sin(t) + a.y * cos(t)); }
PT rotatecw(PT a, double t) { return PT(a.x * cos(t) + a.y * sin(t), -a.x * sin(t) + a.y * cos(t)); }
double SQ(double x) { return x * x; }
double rad_to_deg(double r) { return (r * 180.0 / PI); }
double deg_to_rad(double d) { return (d * PI / 180.0); }
double get_angle(PT a, PT b) {
    double costheta = dot(a, b) / a.norm() / b.norm();
    return acos(max((double)-1.0, min((double)1.0, costheta)));
}
bool is_point_in_angle(PT b, PT a, PT c, PT p) { // does point p lie in angle <bac
    assert(orientation(a, b, c) != 0);
    if (orientation(a, c, b) < 0) swap(b, c);
    return orientation(a, c, p) >= 0 && orientation(a, b, p) <= 0;
}
bool half(PT p) {
    return p.y > 0.0 || (p.y == 0.0 && p.x < 0.0);
}
void polar_sort(vector<PT> &v) { // sort points in counterclockwise
    sort(v.begin(), v.end(), [](PT a,PT b) {
        return make_tuple(half(a), 0.0, a.norm2()) < make_tuple(half(b), cross(a, b), b.norm2());
    });
}
void polar_sort(vector<PT> &v, PT o) { // sort points in counterclockwise with respect to point o
    sort(v.begin(), v.end(), [&](PT a,PT b) {
        return make_tuple(half(a - o), 0.0, (a - o).norm2()) < make_tuple(half(b - o), cross(a - o, b - o), (b - o).norm2());
    });
}
struct line {
    PT a, b; // goes through points a and b
    PT v; double c;  //line form: direction vec [cross] (x, y) = c 
    line() {}
    //direction vector v and offset c
	line(PT v, double c) : v(v), c(c) {
        auto p = get_points();
        a = p.first; b = p.second;
	}
	// equation ax + by + c = 0
	line(double _a, double _b, double _c) : v({_b, -_a}), c(-_c) {
		auto p = get_points();
        a = p.first; b = p.second;
	}
	// goes through points p and q
	line(PT p, PT q) : v(q - p), c(cross(v, p)), a(p), b(q) {}
    	pair<PT, PT> get_points() { //extract any two points from this line
		PT p, q; double a = -v.y, b = v.x; // ax + by = c
		if (sign(a) == 0) {
		    p = PT(0, c / b);
		    q = PT(1, c / b);
		}
		else if (sign(b) == 0) {
		    p = PT(c / a, 0);
		    q = PT(c / a, 1);
		}
		else {
		    p = PT(0, c / b);
		    q = PT(1, (c - a) / b);
		}
		return {p, q};
    	}
    // ax + by + c = 0
    array<double, 3> get_abc() {
        double a = -v.y, b = v.x;
        return {a, b, -c};
    }
    // 1 if on the left, -1 if on the right, 0 if on the line
    int side(PT p) { return sign(cross(v, p) - c); }
    // line that is perpendicular to this and goes through point p
    line perpendicular_through(PT p) { return {p, p + perp(v)}; }
    // translate the line by vector t i.e. shifting it by vector t
    line translate(PT t) { return {v, c + cross(v, t)}; }
    // compare two points by their orthogonal projection on this line
    // a projection point comes before another if it comes first according to vector v
    bool cmp_by_projection(PT p, PT q) { return dot(v, p) < dot(v, q); }
	line shift_left(double d) {
		PT z = v.perp().truncate(d);
		return line(a + z, b + z);
	}
};
// find a point from a through b with distance d
PT point_along_line(PT a, PT b, double d) {
    assert(a != b);
    return a + (((b - a) / (b - a).norm()) * d);
}
// projection point c onto line through a and b  assuming a != b
PT project_from_point_to_line(PT a, PT b, PT c) {
    return a + (b - a) * dot(c - a, b - a) / (b - a).norm2();
}
// reflection point c onto line through a and b  assuming a != b
PT reflection_from_point_to_line(PT a, PT b, PT c) {
    PT p = project_from_point_to_line(a,b,c);
    return p + p - c;
}
// minimum distance from point c to line through a and b
double dist_from_point_to_line(PT a, PT b, PT c) {
    return fabs(cross(b - a, c - a) / (b - a).norm());
}
// returns true if  point p is on line segment ab
bool is_point_on_seg(PT a, PT b, PT p) {
    if (fabs(cross(p - b, a - b)) < eps) {
        if (p.x < min(a.x, b.x) - eps || p.x > max(a.x, b.x) + eps) return false;
        if (p.y < min(a.y, b.y) - eps || p.y > max(a.y, b.y) + eps) return false;
        return true;
    }
    return false;
}
// minimum distance point from point c to segment ab that lies on segment ab
PT project_from_point_to_seg(PT a, PT b, PT c) {
    double r = dist2(a, b);
    if (sign(r) == 0) return a;
    r = dot(c - a, b - a) / r;
    if (r < 0) return a;
    if (r > 1) return b;
    return a + (b - a) * r;
}
// minimum distance from point c to segment ab
double dist_from_point_to_seg(PT a, PT b, PT c) {
    return dist(c, project_from_point_to_seg(a, b, c));
}
// 0 if not parallel, 1 if parallel, 2 if collinear
int is_parallel(PT a, PT b, PT c, PT d) {
    double k = fabs(cross(b - a, d - c));
    if (k < eps){
        if (fabs(cross(a - b, a - c)) < eps && fabs(cross(c - d, c - a)) < eps) return 2;
        else return 1;
    }
    else return 0;
}
// check if two lines are same
bool are_lines_same(PT a, PT b, PT c, PT d) {
    if (fabs(cross(a - c, c - d)) < eps && fabs(cross(b - c, c - d)) < eps) return true;
    return false;
}
// bisector vector of <abc
PT angle_bisector(PT &a, PT &b, PT &c){
    PT p = a - b, q = c - b;
    return p + q * sqrt(dot(p, p) / dot(q, q));
}
// 1 if point is ccw to the line, 2 if point is cw to the line, 3 if point is on the line
int point_line_relation(PT a, PT b, PT p) {
    int c = sign(cross(p - a, b - a));
    if (c < 0) return 1;
    if (c > 0) return 2;
    return 3;
}
// intersection point between ab and cd assuming unique intersection exists
bool line_line_intersection(PT a, PT b, PT c, PT d, PT &ans) {
    double a1 = a.y - b.y, b1 = b.x - a.x, c1 = cross(a, b);
    double a2 = c.y - d.y, b2 = d.x - c.x, c2 = cross(c, d);
    double det = a1 * b2 - a2 * b1;
    if (det == 0) return 0;
    ans = PT((b1 * c2 - b2 * c1) / det, (c1 * a2 - a1 * c2) / det);
    return 1;
}
// intersection point between segment ab and segment cd assuming unique intersection exists
bool seg_seg_intersection(PT a, PT b, PT c, PT d, PT &ans) {
    double oa = cross2(c, d, a), ob = cross2(c, d, b);
    double oc = cross2(a, b, c), od = cross2(a, b, d);
    if (oa * ob < 0 && oc * od < 0){
        ans = (a * ob - b * oa) / (ob - oa);
        return 1;
    }
    else return 0;
}
// intersection point between segment ab and segment cd assuming unique intersection may not exists
// se.size()==0 means no intersection
// se.size()==1 means one intersection
// se.size()==2 means range intersection
set<PT> seg_seg_intersection_inside(PT a,  PT b,  PT c,  PT d) {
    PT ans;
    if (seg_seg_intersection(a, b, c, d, ans)) return {ans};
    set<PT> se;
    if (is_point_on_seg(c, d, a)) se.insert(a);
    if (is_point_on_seg(c, d, b)) se.insert(b);
    if (is_point_on_seg(a, b, c)) se.insert(c);
    if (is_point_on_seg(a, b, d)) se.insert(d);
    return se;
}
// intersection  between segment ab and line cd
// 0 if do not intersect, 1 if proper intersect, 2 if segment intersect
int seg_line_relation(PT a, PT b, PT c, PT d) {
    double p = cross2(c, d, a);
    double q = cross2(c, d, b);
    if (sign(p) == 0 && sign(q) == 0) return 2;
    else if (p * q < 0) return 1;
    else return 0;
}
// intersection between segament ab and line cd assuming unique intersection exists
bool seg_line_intersection(PT a, PT b, PT c, PT d, PT &ans) {
    bool k = seg_line_relation(a, b, c, d);
    assert(k != 2);
    if (k) line_line_intersection(a, b, c, d, ans);
    return k;
}
// minimum distance from segment ab to segment cd
double dist_from_seg_to_seg(PT a, PT b, PT c, PT d) {
    PT dummy;
    if (seg_seg_intersection(a, b, c, d, dummy)) return 0.0;
    else return min({dist_from_point_to_seg(a, b, c), dist_from_point_to_seg(a, b, d), 
        dist_from_point_to_seg(c, d, a), dist_from_point_to_seg(c, d, b)});
}
// minimum distance from point c to ray (starting point a and direction vector b)
double dist_from_point_to_ray(PT a, PT b, PT c) {
    b = a + b;
    double r = dot(c - a, b - a);
    if (r < 0.0) return dist(c, a);
    return dist_from_point_to_line(a, b, c);
}
// starting point as and direction vector ad
bool ray_ray_intersection(PT as, PT ad, PT bs, PT bd) {
    double dx = bs.x - as.x, dy = bs.y - as.y;
    double det = bd.x * ad.y - bd.y * ad.x;
    if (fabs(det) < eps) return 0;
    double u = (dy * bd.x - dx * bd.y) / det;
    double v = (dy * ad.x - dx * ad.y) / det;
    if (sign(u) >= 0 && sign(v) >= 0) return 1;
    else return 0;
}
double ray_ray_distance(PT as, PT ad, PT bs, PT bd) {
    if (ray_ray_intersection(as, ad, bs, bd)) return 0.0;
    double ans = dist_from_point_to_ray(as, ad, bs);
    ans = min(ans, dist_from_point_to_ray(bs, bd, as));
    return ans;
}
struct circle {
    PT p; double r;
    circle() {}
    circle(PT _p, double _r): p(_p), r(_r) {};
    // center (x, y) and radius r
    circle(double x, double y, double _r): p(PT(x, y)), r(_r) {};
    // circumcircle of a triangle
    // the three points must be unique
    circle(PT a, PT b, PT c) {
        b = (a + b) * 0.5;
        c = (a + c) * 0.5;
        line_line_intersection(b, b + rotatecw90(a - b), c, c + rotatecw90(a - c), p);
        r = dist(a, p);
    }
    // inscribed circle of a triangle
    // pass a bool just to differentiate from circumcircle
    circle(PT a, PT b, PT c, bool t) {
        line u, v;
        double m = atan2(b.y - a.y, b.x - a.x), n = atan2(c.y - a.y, c.x - a.x);
        u.a = a;
        u.b = u.a + (PT(cos((n + m)/2.0), sin((n + m)/2.0)));
        v.a = b;
        m = atan2(a.y - b.y, a.x - b.x), n = atan2(c.y - b.y, c.x - b.x);
        v.b = v.a + (PT(cos((n + m)/2.0), sin((n + m)/2.0)));
        line_line_intersection(u.a, u.b, v.a, v.b, p);
        r = dist_from_point_to_seg(a, b, p);
    }
    bool operator == (circle v) { return p == v.p && sign(r - v.r) == 0; }
    double area() { return PI * r * r; }
    double circumference() { return 2.0 * PI * r; }
};
//0 if outside, 1 if on circumference, 2 if inside circle
int circle_point_relation(PT p, double r, PT b) {
    double d = dist(p, b);
    if (sign(d - r) < 0) return 2;
    if (sign(d - r) == 0) return 1;
    return 0;
}
// 0 if outside, 1 if on circumference, 2 if inside circle
int circle_line_relation(PT p, double r, PT a, PT b) {
    double d = dist_from_point_to_line(a, b, p);
    if (sign(d - r) < 0) return 2;
    if (sign(d - r) == 0) return 1;
    return 0;
}
//compute intersection of line through points a and b with
//circle centered at c with radius r > 0
vector<PT> circle_line_intersection(PT c, double r, PT a, PT b) {
    vector<PT> ret;
    b = b - a; a = a - c;
    double A = dot(b, b), B = dot(a, b);
    double C = dot(a, a) - r * r, D = B * B - A * C;
    if (D < -eps) return ret;
    ret.push_back(c + a + b * (-B + sqrt(D + eps)) / A);
    if (D > eps) ret.push_back(c + a + b * (-B - sqrt(D)) / A);
    return ret;
}
//5 - outside and do not intersect
//4 - intersect outside in one point
//3 - intersect in 2 points
//2 - intersect inside in one point
//1 - inside and do not intersect
int circle_circle_relation(PT a, double r, PT b, double R) {
    double d = dist(a, b);
    if (sign(d - r - R) > 0)  return 5;
    if (sign(d - r - R) == 0) return 4;
    double l = fabs(r - R);
    if (sign(d - r - R) < 0 && sign(d - l) > 0) return 3;
    if (sign(d - l) == 0) return 2;
    if (sign(d - l) < 0) return 1;
    assert(0); return -1;
}
vector<PT> circle_circle_intersection(PT a, double r, PT b, double R) {
    if (a == b && sign(r - R) == 0) return {PT(1e18, 1e18)};
    vector<PT> ret;
    double d = sqrt(dist2(a,  b));
    if (d > r + R || d + min(r,  R) < max(r,  R)) return ret;
    double x = (d * d - R * R + r * r) / (2 * d);
    double y = sqrt(r * r - x * x);
    PT v = (b - a) / d;
    ret.push_back(a + v * x  +  rotateccw90(v) * y);
    if (y > 0) ret.push_back(a + v * x - rotateccw90(v) * y);
    return ret;
}
// returns two circle c1, c2 through points a, b and of radius r
// 0 if there is no such circle, 1 if one circle, 2 if two circle
int get_circle(PT a, PT b, double r, circle &c1, circle &c2) {
    vector<PT> v = circle_circle_intersection(a, r, b, r);
    int t = v.size();
    if (!t) return 0;
    c1.p = v[0], c1.r = r;
    if (t == 2) c2.p = v[1], c2.r = r;
    return t;
}
// returns two circle c1, c2 which is tangent to line u,  goes through
// point q and has radius r1; 0 for no circle, 1 if c1 = c2 , 2 if c1 != c2
int get_circle(line u, PT q, double r1, circle &c1, circle &c2) {
    double d = dist_from_point_to_line(u.a, u.b, q);
    if (sign(d - r1 * 2.0) > 0) return 0;
    if (sign(d) == 0) {
    	cout << u.v.x << ' ' << u.v.y << '\n';
        c1.p = q + rotateccw90(u.v).truncate(r1);
        c2.p = q + rotatecw90(u.v).truncate(r1);
        c1.r = c2.r = r1;
        return 2;
    }
    line u1 = line(u.a + rotateccw90(u.v).truncate(r1), u.b + rotateccw90(u.v).truncate(r1));
    line u2 = line(u.a + rotatecw90(u.v).truncate(r1), u.b + rotatecw90(u.v).truncate(r1));
    circle cc = circle(q, r1);
    PT p1, p2; vector<PT> v;
    v = circle_line_intersection(q, r1, u1.a, u1.b);
    if (!v.size()) v = circle_line_intersection(q, r1, u2.a, u2.b);
    v.push_back(v[0]);
    p1 = v[0], p2 = v[1];
    c1 = circle(p1, r1);
    if (p1 == p2) {
        c2 = c1;
        return 1;
    }
    c2 = circle(p2, r1);
    return 2;
}
// returns the circle such that for all points w on the circumference of the circle
// dist(w, a) : dist(w, b) = rp : rq
// rp != rq
// https://e...content-available-to-author-only...a.org/wiki/Circles_of_Apollonius
circle get_apollonius_circle(PT p, PT q, double rp, double rq ){
    rq *= rq ;
    rp *= rp ;
    double a = rq - rp ;
    assert(sign(a));
    double g = rq * p.x - rp * q.x ; g /= a ;
    double h = rq * p.y - rp * q.y ; h /= a ;
    double c = rq * p.x * p.x - rp * q.x * q.x + rq * p.y * p.y - rp * q.y * q.y ;
    c /= a ;
    PT o(g, h);
    double r = g * g + h * h - c ;
    r = sqrt(r);
    return circle(o,r);
}
// returns area of intersection between two circles
double circle_circle_area(PT a, double r1, PT b, double r2) {
    double d = (a - b).norm();
    if(r1 + r2 < d + eps) return 0;
    if(r1 + d < r2 + eps) return PI * r1 * r1;
    if(r2 + d < r1 + eps) return PI * r2 * r2;
    double theta_1 = acos((r1 * r1 + d * d - r2 * r2) / (2 * r1 * d)), 
    	theta_2 = acos((r2 * r2 + d * d - r1 * r1)/(2 * r2 * d));
    return r1 * r1 * (theta_1 - sin(2 * theta_1)/2.) + r2 * r2 * (theta_2 - sin(2 * theta_2)/2.);
}
// tangent lines from point q to the circle
int tangent_lines_from_point(PT p, double r, PT q, line &u, line &v) {
    int x = sign(dist2(p, q) - r * r);
    if (x < 0) return 0; // point in cricle
    if (x == 0) { // point on circle
        u = line(q, q + rotateccw90(q - p));
        v = u;
        return 1;
    }
    double d = dist(p, q);
    double l = r * r / d;
    double h = sqrt(r * r - l * l);
    u = line(q, p + ((q - p).truncate(l) + (rotateccw90(q - p).truncate(h))));
    v = line(q, p + ((q - p).truncate(l) + (rotatecw90(q - p).truncate(h))));
    return 2;
}
// returns outer tangents line of two circles
// if inner == 1 it returns inner tangent lines
int tangents_lines_from_circle(PT c1, double r1, PT c2, double r2, bool inner, line &u, line &v) {
    if (inner) r2 = -r2;
    PT d = c2 - c1;
    double dr = r1 - r2, d2 = d.norm2(), h2 = d2 - dr * dr;
    if (d2 == 0 || h2 < 0) {
        assert(h2 != 0);
        return 0;
    }
    vector<pair<PT, PT>>out;
    for (int tmp: {- 1, 1}) {
        PT v = (d * dr + rotateccw90(d) * sqrt(h2) * tmp) / d2;
        out.push_back({c1 + v * r1, c2 + v * r2});
    }
    u = line(out[0].first, out[0].second);
    if (out.size() == 2) v = line(out[1].first, out[1].second);
    return 1 + (h2 > 0);
}
// O(n^2 log n)
// https://v...content-available-to-author-only...e.net/problem/UVA-12056
struct CircleUnion {
    int n;
    double x[2020], y[2020], r[2020];
    int covered[2020];
    vector<pair<double, double> > seg, cover;
    double arc, pol;
    inline int sign(double x) {return x < -eps ? -1 : x > eps;}
    inline int sign(double x, double y) {return sign(x - y);}
    inline double SQ(const double x) {return x * x;}
    inline double dist(double x1, double y1, double x2, double y2) {return sqrt(SQ(x1 - x2) + SQ(y1 - y2));}
    inline double angle(double A, double B, double C) {
        double val = (SQ(A) + SQ(B) - SQ(C)) / (2 * A * B);
        if (val < -1) val = -1;
        if (val > +1) val = +1;
        return acos(val);
    }
    CircleUnion() {
        n = 0;
        seg.clear(), cover.clear();
        arc = pol = 0;
    }
    void init() {
        n = 0;
        seg.clear(), cover.clear();
        arc = pol = 0;
    }
    void add(double xx, double yy, double rr) {
        x[n] = xx, y[n] = yy, r[n] = rr, covered[n] = 0, n++;
    }
    void getarea(int i, double lef, double rig) {
        arc += 0.5 * r[i] * r[i] * (rig - lef - sin(rig - lef));
        double x1 = x[i] + r[i] * cos(lef), y1 = y[i] + r[i] * sin(lef);
        double x2 = x[i] + r[i] * cos(rig), y2 = y[i] + r[i] * sin(rig);
        pol += x1 * y2 - x2 * y1;
    }
    double solve() {
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < i; j++) {
                if (!sign(x[i] - x[j]) && !sign(y[i] - y[j]) && !sign(r[i] - r[j])) {
                    r[i] = 0.0;
                    break;
                }
            }
        }
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (i != j && sign(r[j] - r[i]) >= 0 && sign(dist(x[i], y[i], x[j], y[j]) - (r[j] - r[i])) <= 0) {
                    covered[i] = 1;
                    break;
                }
            }
        }
        for (int i = 0; i < n; i++) {
            if (sign(r[i]) && !covered[i]) {
                seg.clear();
                for (int j = 0; j < n; j++) {
                    if (i != j) {
                        double d = dist(x[i], y[i], x[j], y[j]);
                        if (sign(d - (r[j] + r[i])) >= 0 || sign(d - abs(r[j] - r[i])) <= 0) {
                            continue;
                        }
                        double alpha = atan2(y[j] - y[i], x[j] - x[i]);
                        double beta = angle(r[i], d, r[j]);
                        pair<double, double> tmp(alpha - beta, alpha + beta);
                        if (sign(tmp.first) <= 0 && sign(tmp.second) <= 0) {
                            seg.push_back(pair<double, double>(2 * PI + tmp.first, 2 * PI + tmp.second));
                        }
                        else if (sign(tmp.first) < 0) {
                            seg.push_back(pair<double, double>(2 * PI + tmp.first, 2 * PI));
                            seg.push_back(pair<double, double>(0, tmp.second));
                        }
                        else {
                            seg.push_back(tmp);
                        }
                    }
                }
                sort(seg.begin(), seg.end());
                double rig = 0;
                for (vector<pair<double, double> >::iterator iter = seg.begin(); iter != seg.end(); iter++) {
                    if (sign(rig - iter->first) >= 0) {
                        rig = max(rig, iter->second);
                    }
                    else {
                        getarea(i, rig, iter->first);
                        rig = iter->second;
                    }
                }
                if (!sign(rig)) {
                    arc += r[i] * r[i] * PI;
                }
                else {
                    getarea(i, rig, 2 * PI);
                }
            }
        }
        return pol / 2.0 + arc;
    }
} CU; 
double area_of_triangle(PT a, PT b, PT c) {
    return fabs(cross(b - a, c - a) * 0.5);
}
// -1 if strictly inside, 0 if on the polygon, 1 if strictly outside
int is_point_in_triangle(PT a, PT b, PT c, PT p) {
    if (sign(cross(b - a,c - a)) < 0) swap(b, c);
    int c1 = sign(cross(b - a,p - a));
    int c2 = sign(cross(c - b,p - b));
    int c3 = sign(cross(a - c,p - c));
    if (c1<0 || c2<0 || c3 < 0) return 1;
    if (c1 + c2 + c3 != 3) return 0;
    return -1;
}
double perimeter(vector<PT> &p) {
    double ans=0; int n = p.size();
    for (int i = 0; i < n; i++) ans += dist(p[i], p[(i + 1) % n]);
    return ans;
}
double area(vector<PT> &p) {
    double ans = 0; int n = p.size();
    for (int i = 0; i < n; i++) ans += cross(p[i], p[(i + 1) % n]);
    return fabs(ans) * 0.5;
}
// centroid of a (possibly non-convex) polygon, 
// assuming that the coordinates are listed in a clockwise or
// counterclockwise fashion.  Note that the centroid is often known as
// the "center of gravity" or "center of mass".
PT centroid(vector<PT> &p) {
    int n = p.size(); PT c(0, 0);
    double sum = 0;
    for (int i = 0; i < n; i++) sum += cross(p[i], p[(i + 1) % n]);
    double scale = 3.0 * sum;
    for (int i = 0; i < n; i++) {
        int j = (i + 1) % n;
        c = c + (p[i] + p[j]) * cross(p[i], p[j]);
    }
    return c / scale;
}
// 0 if cw, 1 if ccw
bool get_direction(vector<PT> &p) {
    double ans = 0; int n = p.size();
    for (int i = 0; i < n; i++) ans += cross(p[i], p[(i + 1) % n]);
    if (sign(ans) > 0) return 1;
    return 0;
}
// it returns a point such that the sum of distances
// from that point to all points in p  is minimum
// O(n log^2 MX)
PT geometric_median(vector<PT> p) {
	auto tot_dist = [&](PT z) {
	    double res = 0;
	    for (int i = 0; i < p.size(); i++) res += dist(p[i], z);
	    return res;
	};
	auto findY = [&](double x) {
	    double yl = -1e5, yr = 1e5;
	    for (int i = 0; i < 60; i++) {
	        double ym1 = yl + (yr - yl) / 3;
	        double ym2 = yr - (yr - yl) / 3;
	        double d1 = tot_dist(PT(x, ym1));
	        double d2 = tot_dist(PT(x, ym2));
	        if (d1 < d2) yr = ym2;
	        else yl = ym1;
	    }
	    return pair<double, double> (yl, tot_dist(PT(x, yl)));
	};
    double xl = -1e5, xr = 1e5;
    for (int i = 0; i < 60; i++) {
        double xm1 = xl + (xr - xl) / 3;
        double xm2 = xr - (xr - xl) / 3;
        double y1, d1, y2, d2;
        auto z = findY(xm1); y1 = z.first; d1 = z.second;
        z = findY(xm2); y2 = z.first; d2 = z.second;
        if (d1 < d2) xr = xm2;
        else xl = xm1;
    }
    return {xl, findY(xl).first };
}
vector<PT> convex_hull(vector<PT> &p) {
	if (p.size() <= 1) return p;
	vector<PT> v = p;
    sort(v.begin(), v.end());
    vector<PT> up, dn;
    for (auto& p : v) {
        while (up.size() > 1 && orientation(up[up.size() - 2], up.back(), p) >= 0) {
            up.pop_back();
        }
        while (dn.size() > 1 && orientation(dn[dn.size() - 2], dn.back(), p) <= 0) {
            dn.pop_back();
        }
        up.push_back(p);
        dn.push_back(p);
    }
    v = dn;
    if (v.size() > 1) v.pop_back();
    reverse(up.begin(), up.end());
    up.pop_back();
    for (auto& p : up) {
        v.push_back(p);
    }
    if (v.size() == 2 && v[0] == v[1]) v.pop_back();
    return v;
}
 //checks if convex or not
bool is_convex(vector<PT> &p) {
    bool s[3]; s[0] = s[1] = s[2] = 0;
    int n = p.size();
    for (int i = 0; i < n; i++) {
        int j = (i + 1) % n;
        int k = (j + 1) % n;
        s[sign(cross(p[j] - p[i], p[k] - p[i])) + 1] = 1;
        if (s[0] && s[2]) return 0;
    }
    return 1;
}
// -1 if strictly inside, 0 if on the polygon, 1 if strictly outside
// it must be strictly convex, otherwise make it strictly convex first
int is_point_in_convex(vector<PT> &p, const PT& x) { // O(log n)
    int n = p.size(); assert(n >= 3);
    int a = orientation(p[0], p[1], x), b = orientation(p[0], p[n - 1], x);
    if (a < 0 || b > 0) return 1;
    int l = 1, r = n - 1;
    while (l + 1 < r) {
        int mid = l + r >> 1;
        if (orientation(p[0], p[mid], x) >= 0) l = mid;
        else r = mid;
    }
    int k = orientation(p[l], p[r], x);
    if (k <= 0) return -k;
    if (l == 1 && a == 0) return 0;
    if (r == n - 1 && b == 0) return 0;
    return -1;
}
bool is_point_on_polygon(vector<PT> &p, const PT& z) {
    int n = p.size();
    for (int i = 0; i < n; i++) {
    	if (is_point_on_seg(p[i], p[(i + 1) % n], z)) return 1;
    }
    return 0;
}
// returns 1e9 if the point is on the polygon 
int winding_number(vector<PT> &p, const PT& z) { // O(n)
    if (is_point_on_polygon(p, z)) return 1e9;
    int n = p.size(), ans = 0;
    for (int i = 0; i < n; ++i) {
        int j = (i + 1) % n;
        bool below = p[i].y < z.y;
        if (below != (p[j].y < z.y)) {
            auto orient = orientation(z, p[j], p[i]);
            if (orient == 0) return 0;
            if (below == (orient > 0)) ans += below ? 1 : -1;
        }
    }
    return ans;
}
// -1 if strictly inside, 0 if on the polygon, 1 if strictly outside
int is_point_in_polygon(vector<PT> &p, const PT& z) { // O(n)
    int k = winding_number(p, z);
    return k == 1e9 ? 0 : k == 0 ? 1 : -1;
}
// id of the vertex having maximum dot product with z
// polygon must need to be convex
// top - upper right vertex
// for minimum dot product negate z and return -dot(z, p[id])
int extreme_vertex(vector<PT> &p, const PT &z, const int top) { // O(log n)
    int n = p.size();
    if (n == 1) return 0;
	double ans = dot(p[0], z); int id = 0;
    if (dot(p[top], z) > ans) ans = dot(p[top], z), id = top;
    int l = 1, r = top - 1;
    while (l < r) {
        int mid = l + r >> 1;
        if (dot(p[mid + 1], z) >= dot(p[mid], z)) l = mid + 1;
        else r = mid;
    }
    if (dot(p[l], z) > ans) ans = dot(p[l], z), id = l;
    l = top + 1, r = n - 1;
    while (l < r) {
        int mid = l + r >> 1;
        if (dot(p[(mid + 1) % n], z) >= dot(p[mid], z)) l = mid + 1;
        else r = mid;
    }
    l %= n;
    if (dot(p[l], z) > ans) ans = dot(p[l], z), id = l;
    return id;
}
// maximum distance from any point on the perimeter to another point on the perimeter
double diameter(vector<PT> &p) {
    int n = (int)p.size();
    if (n == 1) return 0;
    if (n == 2) return dist(p[0], p[1]);
    double ans = 0;
    int i = 0, j = 1;
    while (i < n) {
        while (cross(p[(i + 1) % n] - p[i], p[(j + 1) % n] - p[j]) >= 0) {
        	ans = max(ans, dist2(p[i], p[j]));
        	j = (j + 1) % n;
        }
        ans = max(ans, dist2(p[i], p[j]));
        i++;
    }
    return sqrt(ans);
}
// minimum distance between two parallel lines (non necessarily axis parallel)
// such that the polygon can be put between the lines
double width(vector<PT> &p) {
    int n = (int)p.size();
    if (n <= 2) return 0;
    double ans = inf;
    int i = 0, j = 1;
    while (i < n) {
        while (cross(p[(i + 1) % n] - p[i], p[(j + 1) % n] - p[j]) >= 0) j = (j + 1) % n;
        ans = min(ans, dist_from_point_to_line(p[i], p[(i + 1) % n], p[j]));
        i++;
    }
    return ans;
}
// minimum perimeter
double minimum_enclosing_rectangle(vector<PT> &p) {
	int n = p.size();
	if (n <= 2) return perimeter(p);
	int mndot = 0; double tmp = dot(p[1] - p[0], p[0]);
	for (int i = 1; i < n; i++) {
		if (dot(p[1] - p[0], p[i]) <= tmp) {
			tmp = dot(p[1] - p[0], p[i]);
			mndot = i;
		}
	}
	double ans = inf;
	int i = 0, j = 1, mxdot = 1;
	while (i < n) {
		PT cur = p[(i + 1) % n] - p[i];
        while (cross(cur, p[(j + 1) % n] - p[j]) >= 0) j = (j + 1) % n;
        while (dot(p[(mxdot + 1) % n], cur) >= dot(p[mxdot], cur)) mxdot = (mxdot + 1) % n;
        while (dot(p[(mndot + 1) % n], cur) <= dot(p[mndot], cur)) mndot = (mndot + 1) % n;
        ans = min(ans, 2.0 * ((dot(p[mxdot], cur) / cur.norm() - dot(p[mndot], cur) / cur.norm()) + dist_from_point_to_line(p[i], p[(i + 1) % n], p[j])));
        i++;
    }
    return ans;
}
// given n points, find the minimum enclosing circle of the points
// call convex_hull() before this for faster solution
// expected O(n)
circle minimum_enclosing_circle(vector<PT> &p) {
    random_shuffle(p.begin(), p.end());
    int n = p.size();
    circle c(p[0], 0);
    for (int i = 1; i < n; i++) {
        if (sign(dist(c.p, p[i]) - c.r) > 0) {
            c = circle(p[i], 0);
            for (int j = 0; j < i; j++) {
                if (sign(dist(c.p, p[j]) - c.r) > 0) {
                    c = circle((p[i] + p[j]) / 2, dist(p[i], p[j]) / 2);
                    for (int k = 0; k < j; k++) {
                        if (sign(dist(c.p, p[k]) - c.r) > 0) {
                            c = circle(p[i], p[j], p[k]);
                        }
                    }
                }
            }
        }
    }
    return c;
}
// returns a vector with the vertices of a polygon with everything 
// to the left of the line going from a to b cut away.
vector<PT> cut(vector<PT> &p, PT a, PT b) {
    vector<PT> ans;
    int n = (int)p.size();
    for (int i = 0; i < n; i++) {
        double c1 = cross(b - a, p[i] - a);
        double c2 = cross(b - a, p[(i + 1) % n] - a);
        if (sign(c1) >= 0) ans.push_back(p[i]);
        if (sign(c1 * c2) < 0) {
            if (!is_parallel(p[i], p[(i + 1) % n], a, b)) {
            	PT tmp; line_line_intersection(p[i], p[(i + 1) % n], a, b, tmp);
                ans.push_back(tmp);
            }
        }
    }
    return ans;
}
// not necessarily convex, boundary is included in the intersection
// returns total intersected length
// it returns the sum of the lengths of the portions of the line that are inside the polygon
double polygon_line_intersection(vector<PT> p, PT a, PT b) {
    int n = p.size();
    p.push_back(p[0]);
    line l = line(a, b);
    double ans = 0.0;
    vector< pair<double, int> > vec;
    for (int i = 0; i < n; i++) {
        int s1 = orientation(a, b, p[i]);
        int s2 = orientation(a, b, p[i + 1]);
        if (s1 == s2) continue;
        line t = line(p[i], p[i + 1]);
        PT inter = (t.v * l.c - l.v * t.c) / cross(l.v, t.v);
        double tmp = dot(inter, l.v);
        int f;
        if (s1 > s2) f = s1 && s2 ? 2 : 1;
        else f = s1 && s2 ? -2 : -1;
        vec.push_back(make_pair((f > 0 ? tmp - eps : tmp + eps), f)); // keep eps very small like 1e-12
    }
    sort(vec.begin(), vec.end());
    for (int i = 0, j = 0; i + 1 < (int)vec.size(); i++){
        j += vec[i].second;
        if (j) ans += vec[i + 1].first - vec[i].first; // if this portion is inside the polygon
        // else ans = 0; // if we want the maximum intersected length which is totally inside the polygon, uncomment this and take the maximum of ans
    }
    ans = ans / sqrt(dot(l.v, l.v));
    p.pop_back();
    return ans;
}
// given a convex polygon p, and a line ab and the top vertex of the polygon
// returns the intersection of the line with the polygon
// it returns the indices of the edges of the polygon that are intersected by the line
// so if it returns i, then the line intersects the edge (p[i], p[(i + 1) % n])
array<int, 2> convex_line_intersection(vector<PT> &p, PT a, PT b, int top) {
	int end_a = extreme_vertex(p, (a - b).perp(), top);
    int end_b = extreme_vertex(p, (b - a).perp(), top);
    auto cmp_l = [&](int i) { return orientation(a, p[i], b); };
    if (cmp_l(end_a) < 0 || cmp_l(end_b) > 0)
        return {-1, -1}; // no intersection
    array<int, 2> res;
    for (int i = 0; i < 2; i++) {
        int lo = end_b, hi = end_a, n = p.size();
        while ((lo + 1) % n != hi) {
            int m = ((lo + hi + (lo < hi ? 0 : n)) / 2) % n;
            (cmp_l(m) == cmp_l(end_b) ? lo : hi) = m;
        }
        res[i] = (lo + !cmp_l(hi)) % n;
        swap(end_a, end_b);
    }
    if (res[0] == res[1]) return {res[0], -1}; // touches the vertex res[0]
    if (!cmp_l(res[0]) && !cmp_l(res[1])) 
        switch ((res[0] - res[1] + (int)p.size() + 1) % p.size()) {
            case 0: return {res[0], res[0]}; // touches the edge (res[0], res[0] + 1)
            case 2: return {res[1], res[1]}; // touches the edge (res[1], res[1] + 1)
        }
    return res; // intersects the edges (res[0], res[0] + 1) and (res[1], res[1] + 1)
}
 
pair<PT, int> point_poly_tangent(vector<PT> &p, PT Q, int dir, int l, int r) {
    while (r - l > 1) {
        int mid = (l + r) >> 1;
        bool pvs = orientation(Q, p[mid], p[mid - 1]) != -dir;
        bool nxt = orientation(Q, p[mid], p[mid + 1]) != -dir;
        if (pvs && nxt) return {p[mid], mid};
        if (!(pvs || nxt)) {
            auto p1 = point_poly_tangent(p, Q, dir, mid + 1, r);
            auto p2 = point_poly_tangent(p, Q, dir, l, mid - 1);
            return orientation(Q, p1.first, p2.first) == dir ? p1 : p2;
        }
        if (!pvs) {
            if (orientation(Q, p[mid], p[l]) == dir)  r = mid - 1;
            else if (orientation(Q, p[l], p[r]) == dir) r = mid - 1;
            else l = mid + 1;
        }
        if (!nxt) {
            if (orientation(Q, p[mid], p[l]) == dir)  l = mid + 1;
            else if (orientation(Q, p[l], p[r]) == dir) r = mid - 1;
            else l = mid + 1;
        }
    }
    pair<PT, int> ret = {p[l], l};
    for (int i = l + 1; i <= r; i++) ret = orientation(Q, ret.first, p[i]) != dir ? make_pair(p[i], i) : ret;
    return ret;
}
// (ccw, cw) tangents from a point that is outside this convex polygon
// returns indexes of the points
// ccw means the tangent from Q to that point is in the same direction as the polygon ccw direction
pair<int, int> tangents_from_point_to_polygon(vector<PT> &p, PT Q){
    int ccw = point_poly_tangent(p, Q, 1, 0, (int)p.size() - 1).second;
    int cw = point_poly_tangent(p, Q, -1, 0, (int)p.size() - 1).second;
    return make_pair(ccw, cw);
}
 
// minimum distance from a point to a convex polygon
// it assumes point lie strictly outside the polygon
double dist_from_point_to_polygon(vector<PT> &p, PT z) {
    double ans = inf;
    int n = p.size();
    if (n <= 3) {
        for(int i = 0; i < n; i++) ans = min(ans, dist_from_point_to_seg(p[i], p[(i + 1) % n], z));
        return ans;
    }
    auto [r, l] = tangents_from_point_to_polygon(p, z);
    if(l > r) r += n;
    while (l < r) {
        int mid = (l + r) >> 1;
        double left = dist2(p[mid % n], z), right= dist2(p[(mid + 1) % n], z);
        ans = min({ans, left, right});
        if(left < right) r = mid;
        else l = mid + 1;
    }
    ans = sqrt(ans);
    ans = min(ans, dist_from_point_to_seg(p[l % n], p[(l + 1) % n], z));
    ans = min(ans, dist_from_point_to_seg(p[l % n], p[(l - 1 + n) % n], z));
    return ans;
}
// minimum distance from convex polygon p to line ab
// returns 0 is it intersects with the polygon
// top - upper right vertex
double dist_from_polygon_to_line(vector<PT> &p, PT a, PT b, int top) { //O(log n)
	PT orth = (b - a).perp();
	if (orientation(a, b, p[0]) > 0) orth = (a - b).perp();
	int id = extreme_vertex(p, orth, top);
	if (dot(p[id] - a, orth) > 0) return 0.0; //if orth and a are in the same half of the line, then poly and line intersects
	return dist_from_point_to_line(a, b, p[id]); //does not intersect
}
// minimum distance from a convex polygon to another convex polygon
// the polygon doesnot overlap or touch
// tested in https://t...content-available-to-author-only...h.co/p/the-wall
double dist_from_polygon_to_polygon(vector<PT> &p1, vector<PT> &p2) { // O(n log n)
    double ans = inf;
    for (int i = 0; i < p1.size(); i++) {
        ans = min(ans, dist_from_point_to_polygon(p2, p1[i]));
    }
    for (int i = 0; i < p2.size(); i++) {
        ans = min(ans, dist_from_point_to_polygon(p1, p2[i]));
    }
    return ans;
}
// maximum distance from a convex polygon to another convex polygon
double maximum_dist_from_polygon_to_polygon(vector<PT> &u, vector<PT> &v){ //O(n)
    int n = (int)u.size(), m = (int)v.size();
    double ans = 0;
    if (n < 3 || m < 3) {
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) ans = max(ans, dist2(u[i], v[j]));
        }
        return sqrt(ans);
    }
    if (u[0].x > v[0].x) swap(n, m), swap(u, v);
    int i = 0, j = 0, step = n + m + 10;
    while (j + 1 < m && v[j].x < v[j + 1].x) j++ ;
    while (step--) {
        if (cross(u[(i + 1)%n] - u[i], v[(j + 1)%m] - v[j]) >= 0) j = (j + 1) % m;
        else i = (i + 1) % n;
        ans = max(ans, dist2(u[i], v[j]));
    }
    return sqrt(ans);
}
 
// calculates the area of the union of n polygons (not necessarily convex). 
// the points within each polygon must be given in CCW order.
// complexity: O(N^2), where N is the total number of points
double rat(PT a, PT b, PT p) {
        return !sign(a.x - b.x) ? (p.y - a.y) / (b.y - a.y) : (p.x - a.x) / (b.x - a.x);
 };
double polygon_union(vector<vector<PT>> &p) {
	int n = p.size();
    double ans=0;
    for(int i = 0; i < n; ++i) {
        for (int v = 0; v < (int)p[i].size(); ++v) {
            PT a = p[i][v], b = p[i][(v + 1) % p[i].size()];
            vector<pair<double, int>> segs;
            segs.emplace_back(0,  0), segs.emplace_back(1,  0);
            for(int j = 0; j < n; ++j) {
            	if(i != j) {
                    for(size_t u = 0; u < p[j].size(); ++u) {
                        PT c = p[j][u], d = p[j][(u + 1) % p[j].size()];
                        int sc = sign(cross(b - a, c - a)), sd = sign(cross(b - a, d - a));
                        if(!sc && !sd) {
                            if(sign(dot(b - a, d - c)) > 0 && i > j) {
                                segs.emplace_back(rat(a, b, c), 1), segs.emplace_back(rat(a, b, d),  -1);
                            }
                        } 
                        else {
                            double sa = cross(d - c, a - c), sb = cross(d - c, b - c);
                            if(sc >= 0 && sd < 0) segs.emplace_back(sa / (sa - sb), 1);
                            else if(sc < 0 && sd >= 0) segs.emplace_back(sa / (sa - sb),  -1);
                        }
                    }
                }
            }
            sort(segs.begin(),  segs.end());
            double pre = min(max(segs[0].first, 0.0), 1.0), now, sum = 0;
            int cnt = segs[0].second;
            for(int j = 1; j < segs.size(); ++j) {
                now = min(max(segs[j].first, 0.0), 1.0);
                if (!cnt) sum += now - pre;
                cnt += segs[j].second;
                pre = now;
            }
            ans += cross(a, b) * sum;
        }
    }
    return ans * 0.5;
}
// contains all points p such that: cross(b - a, p - a) >= 0
struct HP {
    PT a, b;
    HP() {}
    HP(PT a, PT b) : a(a), b(b) {}
    HP(const HP& rhs) : a(rhs.a), b(rhs.b) {}
    int operator < (const HP& rhs) const {
        PT p = b - a;
        PT q = rhs.b - rhs.a;
        int fp = (p.y < 0 || (p.y == 0 && p.x < 0));
        int fq = (q.y < 0 || (q.y == 0 && q.x < 0));
        if (fp != fq) return fp == 0;
        if (cross(p, q)) return cross(p, q) > 0;
        return cross(p, rhs.b - a) < 0;
    }
    PT line_line_intersection(PT a, PT b, PT c, PT d) {
        b = b - a; d = c - d; c = c - a;
        return a + b * cross(c, d) / cross(b, d);
    }
    PT intersection(const HP &v) {
        return line_line_intersection(a, b, v.a, v.b);
    }
};
int check(HP a, HP b, HP c) {
    return cross(a.b - a.a, b.intersection(c) - a.a) > -eps; //-eps to include polygons of zero area (straight lines, points)
}
// consider half-plane of counter-clockwise side of each line
// if lines are not bounded add infinity rectangle
// returns a convex polygon, a point can occur multiple times though
// complexity: O(n log(n))
vector<PT> half_plane_intersection(vector<HP> h) {
    sort(h.begin(), h.end());
    vector<HP> tmp;
    for (int i = 0; i < h.size(); i++) {
        if (!i || cross(h[i].b - h[i].a, h[i - 1].b - h[i - 1].a)) {
            tmp.push_back(h[i]);
        }
    }
    h = tmp;
    vector<HP> q(h.size() + 10);
    int qh = 0, qe = 0;
    for (int i = 0; i < h.size(); i++) {
        while (qe - qh > 1 && !check(h[i], q[qe - 2], q[qe - 1])) qe--;
        while (qe - qh > 1 && !check(h[i], q[qh], q[qh + 1])) qh++;
        q[qe++] = h[i];
    }
    while (qe - qh > 2 && !check(q[qh], q[qe - 2], q[qe - 1])) qe--;
    while (qe - qh > 2 && !check(q[qe - 1], q[qh], q[qh + 1])) qh++;
    vector<HP> res; 
    for (int i = qh; i < qe; i++) res.push_back(q[i]);
    vector<PT> hull;
    if (res.size() > 2) {
        for (int i = 0; i < res.size(); i++) {
             hull.push_back(res[i].intersection(res[(i + 1) % ((int)res.size())]));
        }
    }
    return hull;
}
// rotate the polygon such that the (bottom, left)-most point is at the first position
void reorder_polygon(vector<PT> &p) {
  int pos = 0;
  for (int i = 1; i < p.size(); i++) {
    if (p[i].y < p[pos].y || (sign(p[i].y - p[pos].y) == 0 && p[i].x < p[pos].x)) pos = i;
  }
  rotate(p.begin(), p.begin() + pos, p.end());
}
// a and b are convex polygons
// returns a convex hull of their minkowski sum
// min(a.size(), b.size()) >= 2
// https://c...content-available-to-author-only...s.com/geometry/minkowski.html
vector<PT> minkowski_sum(vector<PT> a, vector<PT> b) {
  reorder_polygon(a); reorder_polygon(b);
  int n = a.size(), m = b.size();
  int i = 0, j = 0;
  a.push_back(a[0]); a.push_back(a[1]);
  b.push_back(b[0]); b.push_back(b[1]);
  vector<PT> c;
  while (i < n || j < m) {
    c.push_back(a[i] + b[j]);
    double p = cross(a[i + 1] - a[i], b[j + 1] - b[j]);
    if (sign(p) >= 0) ++i;
    if (sign(p) <= 0) ++j;
  }
  return c;
}
// returns the area of the intersection of the circle with center c and radius r
// and the triangle formed by the points c, a, b
double _triangle_circle_intersection(PT c, double r, PT a, PT b) {
    double sd1 = dist2(c, a), sd2 = dist2(c, b);
    if(sd1 > sd2) swap(a, b), swap(sd1, sd2);
    double sd = dist2(a, b);
    double d1 = sqrtl(sd1), d2 = sqrtl(sd2), d = sqrt(sd);
    double x = abs(sd2 - sd - sd1) / (2 * d);
    double h = sqrtl(sd1 - x * x);
    if(r >= d2) return h * d / 2;
    double area = 0;
    if(sd + sd1 < sd2) {
        if(r < d1) area = r * r * (acos(h / d2) - acos(h / d1)) / 2;
        else {
            area = r * r * ( acos(h / d2) - acos(h / r)) / 2;
            double y = sqrtl(r * r - h * h);
            area += h * (y - x) / 2;
        }
    } 
    else {
        if(r < h) area = r * r * (acos(h / d2) + acos(h / d1)) / 2;
        else {
            area += r * r * (acos(h / d2) - acos(h / r)) / 2;
            double y = sqrtl(r * r - h * h);
            area += h * y / 2;
            if(r < d1) {
                area += r * r * (acos(h / d1) - acos(h / r)) / 2;
                area += h * y / 2;
            } 
            else area += h * x / 2;
        }
    }
    return area;
}
// intersection between a simple polygon and a circle
double polygon_circle_intersection(vector<PT> &v, PT p, double r) {
    int n = v.size();
    double ans = 0.00;
    PT org = {0, 0};
    for(int i = 0; i < n; i++) {
        int x = orientation(p, v[i], v[(i + 1) % n]);
        if(x == 0) continue;
        double area = _triangle_circle_intersection(org, r, v[i] - p, v[(i + 1) % n] - p);
        if (x < 0) ans -= area;
        else ans += area;
    }
    return abs(ans);
}
// find a circle of radius r that contains as many points as possible
// O(n^2 log n);
double maximum_circle_cover(vector<PT> p, double r, circle &c) {
    int n = p.size();
    int ans = 0;
    int id = 0; double th = 0;
    for (int i = 0; i < n; ++i) {
        // maximum circle cover when the circle goes through this point
        vector<pair<double, int>> events = {{-PI, +1}, {PI, -1}};
        for (int j = 0; j < n; ++j) {
            if (j == i) continue;
            double d = dist(p[i], p[j]);
            if (d > r * 2) continue;
            double dir = (p[j] - p[i]).arg();
            double ang = acos(d / 2 / r);
            double st = dir - ang, ed = dir + ang;
            if (st > PI) st -= PI * 2;
            if (st <= -PI) st += PI * 2;
            if (ed > PI) ed -= PI * 2;
            if (ed <= -PI) ed += PI * 2;
            events.push_back({st - eps, +1}); // take care of precisions!
            events.push_back({ed, -1});
            if (st > ed) {
                events.push_back({-PI, +1});
                events.push_back({+PI, -1});
            }
        }
        sort(events.begin(), events.end());
        int cnt = 0;
        for (auto &&e: events) {
            cnt += e.second;
            if (cnt > ans) {
            	ans = cnt;
            	id = i; th = e.first;
            }
        }
    }
    PT w = PT(p[id].x + r * cos(th), p[id].y + r * sin(th));
    c = circle(w, r); //best_circle
    return ans;
}
// radius of the maximum inscribed circle in a convex polygon
double maximum_inscribed_circle(vector<PT> p) {
	int n = p.size();
	if (n <= 2) return 0;
	double l = 0, r = 20000;
	while (r - l > eps) {
		double mid = (l + r) * 0.5;
		vector<HP> h;
		const int L = 1e9;
		h.push_back(HP(PT(-L, -L), PT(L, -L)));
		h.push_back(HP(PT(L, -L), PT(L, L)));
		h.push_back(HP(PT(L, L), PT(-L, L)));
		h.push_back(HP(PT(-L, L), PT(-L, -L)));
		for (int i = 0; i < n; i++) {
			PT z = (p[(i + 1) % n] - p[i]).perp();
			z = z.truncate(mid);
			PT y = p[i] + z, q = p[(i + 1) % n] + z;
			h.push_back(HP(p[i] + z, p[(i + 1) % n] + z));
		}
		vector<PT> nw = half_plane_intersection(h);
		if (!nw.empty()) l = mid;
		else r = mid;
	}
	return l;
}
// ear decomposition, O(n^3) but faster
vector<vector<PT>> triangulate(vector<PT> p) {
  vector<vector<PT>> v;
  while (p.size() >= 3) {
    for (int i = 0, n = p.size(); i < n; i++) {
      int pre = i == 0 ? n - 1 : i - 1;;
      int nxt = i == n - 1 ? 0 : i + 1;;
      int ori = orientation(p[i], p[pre], p[nxt]);
      if (ori < 0) {
        int ok = 1;
        for (int j = 0; j < n; j++) {
          if (j == i || j == pre || j == nxt)continue;
          if (is_point_in_triangle(p[i], p[pre], p[nxt] , p[j]) < 1) {
            ok = 0;
            break;
          }
        }
        if (ok) {
          v.push_back({p[pre], p[i], p[nxt]});
          p.erase(p.begin() + i);
          break;
        }
      }
    }
  }
  return v;
}
 
struct star {
	int n;    // number of sides of the star
	double r; // radius of the circumcircle
	star(int _n, double _r) {
		n = _n;
		r = _r;
	}
 
	double area() {
		double theta = PI / n;
		double s = 2 * r * sin(theta);
		double R = 0.5 * s / tan(theta);
		double a = 0.5 * n * s * R;
		double a2 = 0.25 * s * s / tan(1.5 * theta);
		return a - n * a2;
	}
};
 
// given a list of lengths of the sides of a polygon in counterclockwise order
// returns the maximum area of a non-degenerate polygon that can be formed using those lengths
double get_maximum_polygon_area_for_given_lengths(vector<double> v) {
  if (v.size() < 3) {
    return 0;
  }
  int m = 0;
  double sum = 0;
  for (int i = 0; i < v.size(); i++) {
    if (v[i] > v[m]) {
      m = i;
    }
    sum += v[i];
  }
  if (sign(v[m] - (sum - v[m])) >= 0) {
    return 0; // no non-degenerate polygon is possible
  }
  // the polygon should be a circular polygon
  // that is all points are on the circumference of a circle
  double l = v[m] / 2, r = 1e6; // fix it correctly
  int it = 60;
  auto ang = [](double x, double r) { // x = length of the chord, r = radius of the circle
    return 2 * asin((x / 2) / r);
  };
  auto calc = [=](double r) {
    double sum = 0;
    for (auto x: v) {
      sum += ang(x, r);
    }
    return sum;
  };
  // compute the radius of the circle
  while (it--) {
    double mid = (l + r) / 2;
    if (calc(mid) <= 2 * PI) {
      r = mid;
    }
    else {
      l = mid;
    }
  }
 
  if (calc(r) <= 2 * PI - eps) { // the center of the circle is outside the polygon
    auto calc2 = [&](double r) {
      double sum = 0;
      for (int i = 0; i < v.size(); i++) {
        double x = v[i];
        double th = ang(x, r);
        if (i != m) {
          sum += th;
        }
        else {
          sum += 2 * PI - th;
        }
      }
      return sum;
    };
    l = v[m] / 2; r = 1e6;
    it = 60;
    while (it--) {
      double mid = (l + r) / 2;
      if (calc2(mid) > 2 * PI) {
        r = mid;
      }
      else {
        l = mid;
      }
    }
    auto get_area = [=](double r) {
      double ans = 0;
      for (int i = 0; i < v.size(); i++) {
        double x = v[i];
        double area = r * r * sin(ang(x, r)) / 2;
        if (i != m) {
          ans += area;
        }
        else {
          ans -= area;
        }
      }
      return ans;
    };
    return get_area(r);
  }
  else { // the center of the circle is inside the polygon
    auto get_area = [=](double r) {
      double ans = 0;
      for (auto x: v) {
        ans += r * r * sin(ang(x, r)) / 2;
      }
      return ans;
    };
    return get_area(r);
  }
}
 
int32_t main() {
  	ios_base::sync_with_stdio(0);
  	cin.tie(0);
 
 	return 0;	
}