Ofelia is chased by her evil stepfather, and now she finds herself lost in a labyrinth. She needs your help to run away from her tragic family.

There‘s a huge metal door standing in front of the exit of the labyrinth. There aren dots on the metal door. Pan (the god of the labyrinth) asks Ofelia to find out a triangle which has the largest height. The triangle‘s three vertexes must be the dots on the door, and its area must be positive. Ofelia should tell Pan the triangle‘s height so Pan will let Ofelia go.

Input

There are multiple cases (About 100 cases). For each case, the first line contains an integern (3<=n<=500). In the next n lines, each line contains two real numberx[i], y[i] (0<=x[i], y[i]<=10000) which indicates each dot‘s coordinate. There is no two dots in the same coordinate. The answers are strictly greater than 0.

Output

For each test case, output one line with the maximum triangle height. Any solution with a relative or absolute error of at most 1e-6 will be accepted.

Sample Input

6
0.000 4.000
2.000 4.000
1.000 0.000
1.000 8.000
0.900 7.000
1.100 7.000
7
6967.54555 3457.71200
3.52325 1273.85912
7755.35733 9812.97643
753.00303 2124.70937
7896.71246 8877.78054
5832.77264 5213.70478
4629.38110 8159.01498

Sample Output

7.00000
8940.96643

Hint

In case 1. Choose the dot 3, 5, 6 to make up a triangle, and this triangle has the largest height 7.000.In case 2. We choose dot 2, 3, 5.

#include<iostream>
#include<cstdio>
#include<string>
#include<algorithm>
# include<cmath>
using namespace std;
const int maxn=500+5;
struct point
{
    double x,y;
    point(double x=0,double y=0):x(x),y(y) {}
};

typedef point Vector;


 double cross(Vector A, Vector B)
{
    return A.x*B.y-A.y*B.x;
}

double Dot(Vector A, Vector B)
{
    return A.x*B.x+A.y*B.y;
}
double Length(Vector A)
{
    return sqrt(Dot(A,A));
}
double distanc(point A, point B)
{
    return sqrt((A.x-B.x)*(A.x-B.x)+(A.y-B.y)*(A.y-B.y));
}

Vector operator -(Vector A,Vector B)
{
       return  Vector(A.x-B.x,A.y-B.y);      //被重载坑了一上午。  
}

 double DistanceToLine(point p,point A,point B)
{

    Vector u=B-A,v=p-A;
    return fabs(cross(u,v))/Length(u);
}

/*double dist(point a,point b,point c)
{
    if(b.x!=a.x)
    {
        double k=(b.y-a.y)/(b.x-a.x);
        return fabs((c.y-a.y)-k*(c.x-a.x))/sqrt(1+k*k);
    }
    else
        return fabs(c.x-a.x);
}*/
point num[maxn];
int i,j,k;
int main()
{
    /*#ifndef  ONLINE_JUDGE
    freopen("in.txt","r",stdin);
     #endif*/
    int n;
    while(cin>>n)
    {
        int k;
        double high;
        high=0.0;
      for(i=0;i<n;i++)
        scanf("%lf%lf",&num[i].x,&num[i].y);

      for(i=0;i<n;i++)
       {
            double mmax=0.0;
            for(j=0;j<n;j++)
            {
                if(j==i)  continue;
                if(distanc(num[i],num[j])>mmax)
                {
                       k=j;
                      mmax=distanc(num[i],num[j]);
                }
            }

            for(j=0;j<n;j++)
            {
              if(i==j||j==k)  continue;
              high=max(high,DistanceToLine(num[i],num[j],num[k]));
              high=max(high,DistanceToLine(num[j],num[k],num[i]));
              high=max(high,DistanceToLine(num[k],num[j],num[i]));
             }
       }
       printf("%0.6f\n",high);
    }
    return 0;
}

Pan's Labyrinth （找组成的三角形最大的高）