題幹：

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.

題目大意：

   給你n個點，讓你從這些點裡面尋找4個點，使得四邊形的面積最大。當然題目說了沒有三個點在一條線上，沒有重合的點。

解題報告：

   這題看起來不難啊，，把四個點拆成兩個三角形然後列舉2+1+1，三重迴圈就搞定了，，但是這題還是很坑的啊，，WA23的同黨肯定不少吧，，看這個簡單的B題當時cf比賽的時候比D過題的人都少（不過也可能因為這場的D確實簡單，看懂了就是個模板題）

AC程式碼：

#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;//這麼用還是小心點、、、
//			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
*/

總結：

  sgn函式要用啊，，，double別直接等號判斷，，很危險、、

   這是個坑啊，，還是自己太不小心了、、