Problem:
Score : 600 points

Problem Statement

There are N holes in a two-dimensional plane. The coordinates of the i-th hole are (x_i,y_i).

Let R=10^{10^{10^{10}}}. Ringo performs the following operation:

Randomly choose a point from the interior of a circle of radius R centered at the origin, and put Snuke there. Snuke will move to the hole with the smallest Euclidean distance from the point, and fall into that hole. If there are multiple such holes, the hole with the smallest index will be chosen.

For every i (1 \leq i \leq N), find the probability that Snuke falls into the i-th hole.

Here, the operation of randomly choosing a point from the interior of a circle of radius R is defined as follows:

Pick two real numbers x and y independently according to uniform distribution on [-R,R].

If x^2+y^2\leq R^2, the point (x,y) is chosen. Otherwise, repeat picking the real numbers x,y until the condition is met.

Constraints

2 \leq N \leq 100

|x_i|,|y_i| \leq 10^6(1\leq i\leq N)

All given points are pairwise distinct.

All input values are integers.

Input

Input is given from Standard Input in the following format:

N
x_1 y_1
:
x_N y_N

Output

Print N real numbers. The i-th real number must represent the probability that Snuke falls into the i-th hole.

The output will be judged correct when, for all output values, the absolute or relative error is at most 10^{-5}.

Sample Input 1

2
0 0
1 1

Sample Output 1

0.5
0.5

If Ringo put Snuke in the region x+y\leq 1, Snuke will fall into the first hole. The probability of this happening is very close to 0.5.
Otherwise, Snuke will fall into the second hole, the probability of which happening is also very close to 0.5.

Sample Input 2

5
0 0
2 8
4 5
2 6
3 10

Sample Output 2

0.43160120892732328768
0.03480224363653196956
0.13880483535586193855
0.00000000000000000000
0.39479171208028279727

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define rep(i, n) for(int i = 0; i < n; i++)
#define rep2(i, x, n) for(int i = x; i <= n; i++)
#define rep3(i, x, n) for(int i = x; i >= n; i--)
#define elif else if
#define sp(x) fixed << setprecision(x)
#define pb push_back
#define eb emplace_back
#define all(x) x.begin(), x.end()
#define sz(x) (int)x.size()
using ll = long long;
using ld = long double;
using pii = pair<int, int>;
using pil = pair<int, ll>;
using pli = pair<ll, int>;
using pll = pair<ll, ll>;
const ll MOD = 1e9+7;
//const ll MOD = 998244353;
const int inf = (1<<30)-1;
const ll INF = (1LL<<60)-1;
const ld EPS = 1e-10;
template<typename T> bool chmax(T &x, const T &y) {return (x < y)? (x = y, true) : false;};
template<typename T> bool chmin(T &x, const T &y) {return (x > y)? (x = y, true) : false;};

//点(x, y)
struct P {
    ld x, y;
    P() {}
    P(ld x, ld y) : x(x), y(y) {}

    P &operator += (const P &p) {
        x += p.x, y += p.y;
        return *this;
    }

    P &operator -= (const P &p) {
        x -= p.x, y -= p.y;
        return *this;
    }

    P &operator *= (const ld &d) {
        x *= d, y *= d;
        return *this;
    }

    P &operator /= (const ld &d) {
        x /= d, y /= d;
        return *this;
    }

    P operator + (const P &p) const {return P (*this) += p;}

    P operator - (const P &p) const {return P (*this) -= p;}

    P operator * (const ld &d) const {return P(*this) *= d;}

    P operator / (const ld &d) const {return P(*this) /= d;}

    ld norm() const {return sqrt(x*x+y*y);}

    ld norm2() const {return x*x+y*y;}
};

ld dist(P a, P b) {return (b-a).norm();}

ld dist2(P a, P b){return (b-a).norm2();}

ld dot(P a, P b) {return a.x*b.x+a.y*b.y;}

ld det(P a, P b) {return a.x*b.y-b.x*a.y;}

P mid(P a, P b) {return (a+b)/2;}

//点a,bをs:tに内分する点
P ind(P a, P b, ld s, ld t) {return (a*t+b*s)/(s+t);}

//点a,bをs:tに外分する点
P exd(P a, P b, ld s, ld t) {return ind(a, b, s, -t);}

//直線ax+by+c=0
struct L {ld a, b, c;};

//(x, y)を中心とする半径rの円
struct C {ld x, y, r;};

//直線l1と直線l2の交点
vector<P> l_l(L l1, L l2){
    vector<P> res;
    if(abs(l1.a*l2.b - l2.a*l1.b) < EPS) return res;
    ld x = (l2.c*l1.b-l1.c*l2.b)/(l1.a*l2.b-l2.a*l1.b);
    ld y = (l2.c*l1.a-l1.c*l2.a)/(l1.b*l2.a-l2.b*l1.a);
    res.pb(P(x, y));
    return res;
}

//直線lと円cの交点
vector<P> l_c(L l, C c){
    vector<P> res;
    if(l.a == 0 && l.b == 0) return res;
    //直線と円の中心の距離
    ld d = abs(l.a*c.x+l.b*c.y+l.c)/sqrt(l.a*l.a+l.b*l.b);
    if(d-c.r > EPS) return res;
    //hx,hy:=円の中心から直線に下ろした垂線の足の座標
    ld hx = -(l.a*l.c+l.b*(l.a*c.y-l.b*c.x))/(l.a*l.a+l.b*l.b);
    ld hy = -(l.b*l.c+l.a*(l.a*c.y-l.b*c.x))/(l.a*l.a+l.b*l.b);
    if(abs(d-c.r) < EPS){
        res.pb(P(hx, hy));
        return res;
    }
    //d2:=垂線の足hと交点の距離
    ld d2 = sqrt(c.r*c.r-d*d);
    //ex,ey:=直線と同じ向きの単位ベクトル
    ld ex = -l.b/sqrt(l.a*l.a+l.b*l.b), ey = l.a/sqrt(l.a*l.a+l.b*l.b);
    res.pb(P(hx+d2*ex, hy+d2*ey)), res.pb(P(hx-d2*ex, hy-d2*ey));
    return res;
}

//円c1と円c2の交点
vector<P> c_c(C c1, C c2){
    //直線ax+by+c=0:=２つの円の交点を通る直線
    ld a = 2*(c1.x-c2.x), b = 2*(c1.y-c2.y);
    ld c = -(c1.x*c1.x+c1.y*c1.y-c1.r*c1.r)+(c2.x*c2.x+c2.y*c2.y-c2.r*c2.r);
    L l = {a, b, c};
    return l_c(l, c1);
}

//点a,bを通る直線
L make_line(P a, P b){
    ld p = a.x, q = a.y, r = b.x, s = b.y;
    ld A = s-q, B = p-r, C = -p*s+q*r;
    L l = {A, B, C};
    return l;
}

//点a,bの垂直二等分線
L vertical_bisector(P a, P b){
    ld p = a.x, q = a.y, r = b.x, s = b.y;
    ld A = r-p, B = s-q, C = (p*p-r*r+q*q-s*s)/2;
    L l = {A, B, C};
    return l;
}

//点aから直線lに降ろした垂線の足
P foot(P a, L l){
    ld A = l.b, B = -l.a, C = l.a*a.y - l.b*a.x;
    L subl = {A, B, C};
    vector<P> p = l_l(l, subl);
    return p[0];
}

//点aがp1とp2を結ぶ線分上にあるかどうか
bool on_line(P a, P p1, P p2){
    P sub1 = p1-a, sub2 = p2-a;
    return (abs(det(sub1, sub2)) < EPS && dot(sub1, sub2) < EPS);
}

//x座標、y座標の小さい順にソート
bool comp_x(const P &p, const P &q){
    if(p.x != q.x) return p.x < q.x;
    return p.y < q.y;
}

//点の集合psの凸包を求める
vector<P> convex_hull(vector<P> ps){
    sort(all(ps), comp_x);
    int n = ps.size(), k = 0;
    vector<P> qs(2*n);
    //上側凸包を作成
    rep(i, n){
        while(k > 1 && det(qs[k-2]-qs[k-1], ps[i]-qs[k-1]) < EPS) k--;
        qs[k++] = ps[i];
    }
    int t = k;
    //下側凸包を作成
    rep3(i, n-2, 0){
        while(k > t && det(qs[k-2]-qs[k-1], ps[i]-qs[k-1]) < EPS) k--;
        qs[k++] = ps[i];
    }
    qs.resize(k-1);
    return qs;
}

int main(){
    ld pi = 3.14159265358979;
    int N;
    cin >> N;
    vector<P> ps;
    rep(i, N){
        ld x, y; cin >> x >> y;
        ps.pb(P(x, y));
    }
    vector<P> ch = convex_hull(ps);
    cout << sp(10);
    int n = sz(ch);
    if(n == 2){
        rep(i, N){
            if(ps[i].x == ch[0].x && ps[i].y == ch[0].y) cout << 0.5 << endl;
            elif(ps[i].x == ch[1].x && ps[i].y == ch[1].y) cout << 0.5 << endl;
            else cout << 0 << endl;
        }
        return 0;
    }
    ld ans[N];
    rep(i, N) ans[i] = 0;
    rep(j, n){
        int i = (j+n-1)%n, k = (j+1)%n;
        P a = ch[i]-ch[j], b = ch[k]-ch[j];
        ld cos = dot(a, b)/(a.norm()*b.norm());
        ld the = pi-acos(cos);
        rep(l, N){
            if(ps[l].x == ch[j].x && ps[l].y == ch[j].y) ans[l] = the/(2*pi); 
        }
    }
    rep(i, N) cout << ans[i] << endl;
}