题干：

Iahub has drawn a set of n points in the cartesian plane which he calls "special points". A quadrilateral is a simple polygon without self-intersections with four sides (also called edges) and four vertices (also called corners). Please note that a quadrilateral doesn't have to be convex. A special quadrilateral is one which has all four vertices in the set of special points. Given the set of special points, please calculate the maximal area of a special quadrilateral.

Input

The first line contains integer n (4 ≤ n ≤ 300). Each of the next n lines contains two integers: xi, yi ( - 1000 ≤ xi, yi ≤ 1000) — the cartesian coordinates of ith special point. It is guaranteed that no three points are on the same line. It is guaranteed that no two points coincide.

Output

Output a single real number — the maximal area of a special quadrilateral. The answer will be considered correct if its absolute or relative error does't exceed 10 - 9.

Examples

Input

5
0 0
0 4
4 0
4 4
2 3

Output

16.000000

Note

In the test example we can choose first 4 points to be the vertices of the quadrilateral. They form a square by side 4, so the area is 4·4 = 16.

题目大意：

   给你n个点，让你从这些点里面寻找4个点，使得四边形的面积最大。当然题目说了没有三个点在一条线上，没有重合的点。

解题报告：

   这题看起来不难啊，，把四个点拆成两个三角形然后枚举2+1+1，三重循环就搞定了，，但是这题还是很坑的啊，，WA23的同党肯定不少吧，，看这个简单的B题当时cf比赛的时候比D过题的人都少（不过也可能因为这场的D确实简单，看懂了就是个模板题）

AC代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define ll long long
#define pb push_back
#define pm make_pair
#define fi first
#define se second
using namespace std;
const int MAX = 2e5 + 5;
const double eps = 1e-8;
int sgn(double x) {if(fabs(x) < eps)return 0;if(x < 0) return -1;return 1;
}
struct Point {double x,y;int id;Point() {}Point(double x,double y):x(x),y(y) {}Point operator -(const Point &b)const {return Point(x - b.x,y - b.y);}double operator ^(const Point &b)const {return x*b.y - y*b.x;}double operator *(const Point &b)const {return x*b.x + y*b.y;}
} p[1005];
double ans,ans1,ans2;
int main()
{int n;cin>>n;for(int i = 1; i<=n; i++) scanf("%lf%lf",&p[i].x,&p[i].y),p[i].id = i;for(int i = 1; i<=n; i++) {for(int j = 1; j<=n; j++) {ans1=ans2=-1.0;for(int k = 1; k<=n; k++) {if(k == i || k == j) continue;double tmp = (p[k]-p[i])^(p[j]-p[i]);	if(sgn(tmp) > 0) ans1 = max(ans1,tmp);if(sgn(tmp) < 0) ans2 = max(ans2,-tmp);}if(ans1==-1.0 || ans2 == -1.0) continue;//这么用还是小心点、、、
//			cout << i << "***"<<j<< "***";
//			cout << ans1+ans2<<endl;ans = max(ans,ans1+ans2);}}printf("%.8f\n",ans/2);return 0 ;}
/*
4
0 0
0 5
5 0
1 1
*/

总结：

  sgn函数要用啊，，，double别直接等号判断，，很危险、、

   这是个坑啊，，还是自己太不小心了、、