codeforces Volatile Kite(几何）

D. Volatile Kite

time limit per test

2 seconds

memory limit per test

256 megabytes

You are given a convex polygon P with n distinct vertices p₁, p₂, ..., p_n. Vertex p_i has coordinates (x_i, y_i) in the 2D plane. These vertices are listed in clockwise order.

You can choose a real number D and move each vertex of the polygon a distance of at mostD from their original positions.

Find the maximum value of D such that no matter how you move the vertices, the polygon does not intersect itself and stays convex.

Input

The first line has one integer n (4 ≤ n ≤ 1 000) — the number of vertices.

The next n lines contain the coordinates of the vertices. Linei contains two integersx_i andy_i ( - 10⁹ ≤ x_i, y_i ≤ 10⁹) — the coordinates of the i-th vertex. These points are guaranteed to be given in clockwise order, and will form a strictly convex polygon (in particular, no three consecutive points lie on the same straight line).

Output

Print one real number D, which is the maximum real number such that no matter how you move the vertices, the polygon stays convex.

Your answer will be considered correct if its absolute or relative error does not exceed10^- 6.

Namely, let's assume that your answer is a and the answer of the jury isb. The checker program will consider your answer correct if.

Examples

Input

4
0 0
0 1
1 1
1 0

Output

0.3535533906

Input

6
5 0
10 0
12 -4
10 -8
5 -8
3 -4

Output

1.0000000000

Note

Here is a picture of the first sample

Here is an example of making the polygon non-convex.

This is not an optimal solution, since the maximum distance we moved one point is ≈ 0.4242640687, whereas we can make it non-convex by only moving each point a distance of at most ≈ 0.3535533906.

题意：给你一个凸多边形，问至少移动多少距离（每个点都可以移动）使得不是凸多边形；

思路：对相邻的三个点来说，选定中间点，使得三个点同时向中间移动直到共线就好。

代码：

#include
using namespace std;
const int maxn=1e5+100;
struct Point
{
    double x,y;
    Point(double _x=0,double _y=0)
    {
        x=_x;
        y=_y;
    }


} point[maxn];
Point operator-( const Point & a, const Point&b)
{
    return Point(a.x-b.x,a.y-b.y);
}
Point operator+(const Point&a,const Point&b)
{
    return Point(a.x+b.x,a.y+b.y);
}
double operator*(const Point&a,const Point&b)//向量积的数值，等于两个向量围成的平行四边形的面积。
{
    return a.x*b.y-b.x*a.y;
}
double dis(const Point&a,const Point&b)
{
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=0; i