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>
const int MAXN = 300;
int x[MAXN], y[MAXN];
int main() {
  int n;
  int i, j, k;
  int ax, ay;
  int t, min, max, ans;
  scanf("%d", &n);
  for (i = 0; i < n; ++i) scanf("%d%d", &x[i], &y[i]);
  ans = 0;
  for (i = 0; i < n; ++i)
    for (j = i + 1; j < n; ++j) {
      min = max = 0;
      ax = x[j] - x[i];
      ay = y[j] - y[i];
      for (k = 0; k < n; ++k) {
        t = ax * (y[k] - y[i]) - ay * (x[k] - x[i]);
        if (t < min)
          min = t;
        else if (t > max)
          max = t;
      }
      if (min < 0 && max > 0 && max - min > ans) ans = max - min;
    }
  printf("%d", ans >> 1);
  if (ans & 1) printf(".5");
  return 0;
}