You are given a convex polygon P with n distinct
vertices p1, p2, ..., pn.
Vertex pi has
coordinates (xi, yi) 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 most D 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.

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. Line i contains
two integers xi and yi ( - 109 ≤ xi, yi ≤ 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).

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 exceed 10 - 6.

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

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.

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题目的意思是按顺序给出n个点坐标，任意移动每个点，使得凸多边形变成凹多边形，

问最大移动的点最少移动多少

观察我们很容易发现，只要把相邻3个点中间的点向另外两个点构成的直线垂直移动，

那两个点向外移动一定最小，枚举所有点即可

#include <iostream>

#include <cstdio>

#include <cstring>

#include <cmath>

#include <algorithm>

#include <string>

#include <set>

#include <map>

#include <queue>

using namespace std;

#define inf 0x3f3f3f3f

struct point

{

    double x,y;

};

point intersection(point u1,point u2,point v1,point v2)

{

    point ret=u1;

    double t=((u1.x-v1.x)*(v1.y-v2.y)-(u1.y-v1.y)*(v1.x-v2.x))

            /((u1.x-u2.x)*(v1.y-v2.y)-(u1.y-u2.y)*(v1.x-v2.x));

    ret.x+=(u2.x-u1.x)*t;

    ret.y+=(u2.y-u1.y)*t;

    return ret;

}

point ptoline(point p,point l1,point l2)

{

    point t=p;

    t.x+=l1.y-l2.y,t.y+=l2.x-l1.x;

    return intersection(p,t,l1,l2);

}

double dis(point p1,point p2)

{

    return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y));

}

int main()

{

    point p[1006];

    int n;

    while(~scanf("%d",&n))

    {

        for(int i=0; i<n; i++)

        {

            scanf("%lf%lf",&p[i].x,&p[i].y);

        }

        p[n]=p[0];

        p[n+1]=p[1];

        double ans=inf;

        for(int i=0; i<n; i++)

        {

            double x=dis(p[i+1],ptoline(p[i+1],p[i],p[i+2]))/2;

            ans=min(ans,x);

        }

        printf("%.10lf\n",ans);

    }

    return 0;

}