Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
double Check(int x1, int y1, int x2, int y2, int x3, int y3) {
  double Ans = (double)((x2 - x1) * (y3 - y2) - (y2 - y1) * (x3 - x2));
  return (double)Ans / 2.00000;
}
int main() {
  int N;
  cin >> N;
  int a[N + 5][2];
  int x, y;
  for (int n = 0; n < N; ++n) {
    scanf("%d%d", &x, &y);
    a[n][0] = x;
    a[n][1] = y;
  }
  double Ans = 0;
  int temp;
  for (int i = 0; i < N; ++i) {
    for (int j = i + 1; j < N; ++j) {
      double Ans1 = -1;
      double Ans2 = -1;
      for (int k = 0; k < N; ++k) {
        if (k != i && k != j) {
          if (Check(a[i][0], a[i][1], a[j][0], a[j][1], a[k][0], a[k][1]) < 0) {
            Ans1 = max(Ans1, -Check(a[i][0], a[i][1], a[j][0], a[j][1], a[k][0],
                                    a[k][1]));
          } else {
            Ans2 = max(Ans2, Check(a[i][0], a[i][1], a[j][0], a[j][1], a[k][0],
                                   a[k][1]));
          }
        }
      }
      if (Ans1 >= 0 && Ans2 >= 0) {
        Ans = max(Ans, (double)Ans1 + Ans2);
      }
    }
  }
  printf("%.9lf\n", (double)Ans);
  return 0;
}