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;
const int maxn = 305;
int x[maxn], y[maxn];
double calc(int i, int j, int k) {
  return ((x[j] - x[i]) * (y[k] - y[j]) - (y[j] - y[i]) * (x[k] - x[j])) / 2.0;
}
int main() {
  ios::sync_with_stdio(false);
  int n;
  double ans = 0.0;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%d %d", &x[i], &y[i]);
  for (int i = 0; i < n; ++i) {
    for (int j = i + 1; j < n; ++j) {
      double mx1 = -1, mx2 = -1, tmp;
      for (int k = 0; k < n; ++k) {
        if (k == i || k == j) continue;
        tmp = calc(i, j, k);
        if (tmp < 0)
          mx1 = max(mx1, -tmp);
        else
          mx2 = max(mx2, tmp);
      }
      if (mx1 >= 0 && mx2 >= 0) ans = max(ans, mx1 + mx2);
    }
  }
  printf("%.12lf\n", ans);
  return 0;
}