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 point {
  int x, y;
  point(int x = 0, int y = 0) : x(x), y(y) {}
  point operator-(point other) const { return point(x - other.x, y - other.y); }
};
point pt[311];
int Cross(point v1, point v2) { return v1.x * v2.y - v1.y * v2.x; }
int main() {
  int n, i, j, k;
  scanf("%d", &n);
  for (i = 1; i <= n; ++i) {
    scanf("%d%d", &pt[i].x, &pt[i].y);
  }
  int ans = 0;
  for (i = 1; i <= n; ++i) {
    for (j = i + 1; j <= n; ++j) {
      point diagnal = pt[j] - pt[i];
      int mx = 0, mn = 0;
      for (k = 1; k <= n; ++k) {
        if (i == k || j == k) continue;
        point vc = pt[k] - pt[i];
        int area = Cross(diagnal, vc);
        if (mn == 0 || mn > area) mn = area;
        if (mx == 0 || mx < area) mx = area;
      }
      if (mx != 0 && mn != 0) {
        ans = max(mx - mn, ans);
      }
    }
  }
  printf("%lf\n", 0.5 * ans);
  return 0;
}