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;
};
struct triangle {
  point a, b, c;
  long double area() {
    long double A = a.x * (b.y - c.y) + b.x * (c.y - a.y) + c.x * (a.y - b.y);
    if (A < 0) A = A * -1.0;
    return A / 2.0;
  }
  bool ccw() {
    return (b.x - a.x) * (c.y - a.y) - (c.x - a.x) * (b.y - a.y) > 0;
  }
};
int n;
point p[333];
long double max_quadrilateral_area() {
  long double max_area = 0;
  for (int i = 0; i < n - 1; i++)
    for (int j = i + 1; j < n; j++) {
      long double max_ccw = 0, max_cw = 0;
      for (int k = 0; k < n; k++)
        if (k != i && k != j) {
          triangle t{p[i], p[j], p[k]};
          if (t.ccw())
            max_ccw = max(max_ccw, t.area());
          else
            max_cw = max(max_cw, t.area());
        }
      if (max_ccw > 0 && max_cw > 0) max_area = max(max_area, max_ccw + max_cw);
    }
  return max_area;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d%d", &p[i].x, &p[i].y);
  }
  printf("%Lf\n", max_quadrilateral_area());
  return 0;
}