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 PT {
  int x, y;
  PT() {}
  PT(int x, int y) : x(x), y(y) {}
  PT operator+(const PT &p) const { return PT(x + p.x, y + p.y); }
  PT operator-(const PT &p) const { return PT(x - p.x, y - p.y); }
  int operator%(const PT &p) const { return x * p.y - y * p.x; }
};
PT arr[305];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = (0); i < (n); ++i) scanf("%d %d", &arr[i].x, &arr[i].y);
  int ans = 0;
  for (int i = (0); i < (n); ++i)
    for (int j = (i + 1); j < (n); ++j) {
      int left = 0, right = 0;
      bool ok1 = false, ok2 = false;
      for (int k = (0); k < (n); ++k) {
        if (k != i && k != j) {
          int area = (arr[k] - arr[i]) % (arr[j] - arr[i]);
          if (area < 0)
            left = max(left, -area), ok1 = true;
          else
            right = max(right, area), ok2 = true;
        }
      }
      if (ok1 && ok2) ans = max(ans, right + left);
    }
  printf("%lf\n", (double)ans / 2.0);
  return 0;
}