Problem: 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}.

Examples

Input

2
0 0
1 1


Output

0.5
0.5


Input

5
0 0
2 8
4 5
2 6
3 10


Output

0.43160120892732328768
0.03480224363653196956
0.13880483535586193855
0.00000000000000000000
0.39479171208028279727

Solution (in CPP):

Solution:
//点iの領域は, 線分i-jの垂直2等分線(j≠i, 1≦j≦n)によって作られるが、全体を1としたときの面積は垂直2等分線同士の角度で求められる。
//変形すると、単に方向ベクトルi->j同士(i:固定, j:i以外)の角度で求められる。
//ある方向ベクトルi->j1を基準(Re=1, Im=0)に取り、反時計回りに進む方向を角度(0～2 * PI)正としたときの、方向ベクトルの最大角度
//をA(i, j1)とおくと、X = min(A(i, 1), …, A(i, n)) (2項目はi以外)がPI以上なら「i行目の答え = 0」、
//PI未満なら「i行目の答え = (PI - X) / (2 * PI)」になる。「基準i->j1以外の方向ベクトルの角度値を0超えにしたいこと・第3,4象限」を考慮すると、
//円環数列のテクの応用によって、O(N^2)に落とせる。
//注意点としては、円環数列を作る前に角度をソートする必要があること！！！

#include <iostream>
#include <complex>
#include <algorithm>
#include <cstdio>
using namespace std;
typedef complex<double> P;

double PI = 3.14159265358979;
int n;
P p[100];

int main() {
	int i, j, k;
	cin >> n;
	for (i = 0; i < n; i++) {
		double x, y; cin >> x >> y;
		p[i] = P(x, y);
	}
	
	for (i = 0; i < n; i++) {
		double ang[198];
		int cnt = 0;
		for (j = 0; j < n; j++) {
			if (j == i) continue;
			ang[cnt++] = arg(p[j] - p[i]);
		}
		sort(ang, ang + cnt);
		for (j = 0; j < cnt; j++) {
			ang[cnt + j] = ang[j] + 2 * PI;
		}
		
		double minArg = PI;
		for (j = 0; j < cnt; j++) {
			minArg = min(minArg, ang[j + cnt - 1] - ang[j]);
		}
		double ans = (PI - minArg) / (2 * PI);
		printf("%.14f\n", ans);
	}
	return 0;
}