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;
struct node {
  double x, y;
} a[500];
double sum(node a, node b, node c) {
  return 0.5 * ((b.x - c.x) * (a.y - c.y) - (a.x - c.x) * (b.y - c.y));
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%lf%lf", &a[i].x, &a[i].y);
  }
  double ans = 0, ans1 = 0, ans2 = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = i + 1; j <= n; j++) {
      ans2 = 0, ans1 = 0;
      for (int k = 1; k <= n; k++) {
        if (k == i || k == j) continue;
        double tmp = sum(a[i], a[j], a[k]);
        if (tmp > 0)
          ans1 = max(ans1, tmp);
        else if (tmp < 0)
          ans2 = max(ans2, -tmp);
      }
      if (ans1 > 0 && ans2 > 0) ans = max(ans, ans1 + ans2);
    }
  }
  printf("%lf\n", ans);
}