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;
  int y;
} point[305], p1, p2;
int main() {
  int n;
  scanf("%d", &n);
  double ans = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d %d", &point[i].x, &point[i].y);
  }
  for (int i = 1; i <= n - 1; i++) {
    for (int j = i + 1; j <= n; j++) {
      p1.x = point[j].x - point[i].x;
      p1.y = point[j].y - point[i].y;
      double d1 = 0, d2 = 0;
      double side = sqrt(p1.x * p1.x * 1.0 + p1.y * p1.y * 1.0);
      for (int k = 1; k <= n; k++) {
        if (k == i || k == j) continue;
        p2.x = point[k].x - point[i].x;
        p2.y = point[k].y - point[i].y;
        double dd = p1.x * p2.y - p2.x * p1.y;
        double t = fabs(dd / side);
        if (dd < 0) {
          if (d1 < t) d1 = t;
        } else {
          if (d2 < t) d2 = t;
        }
      }
      if (d1 != 0 && d2 != 0) {
        double area = 0.5 * side * (d1 + d2);
        if (area > ans) ans = area;
      }
    }
  }
  printf("%.10f", ans);
  return 0;
}