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>
struct Coor {
  double x, y;
} p[301];
int sgn(double x) { return x < 1e-6 && x > -1e-6 ? 0 : x > 0 ? 1 : -1; }
double area_sgn(Coor a, Coor b, Coor c) {
  return a.x * b.y + b.x * c.y + c.x * a.y - b.x * a.y - c.x * b.y - a.x * c.y;
}
int dir(Coor a, Coor b, Coor c) { return sgn(area_sgn(a, b, c)); }
int main() {
  int i, j, k;
  int n;
  int p1, p2;
  double ans = 0;
  double max1, max2;
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%lf%lf", &p[i].x, &p[i].y);
  }
  for (p1 = 0; p1 < n; p1++) {
    for (p2 = p1 + 1; p2 < n; p2++) {
      max1 = max2 = 0;
      for (k = 0; k < n; k++) {
        if (k == p1 || k == p2) continue;
        double area = area_sgn(p[k], p[p1], p[p2]);
        if (area > 0) {
          if (max1 < area) max1 = area;
        } else if (area < 0) {
          if (max2 < -area) max2 = -area;
        }
      }
      if (ans < max1 + max2 && max1 != 0 && max2 != 0) ans = max1 + max2;
    }
  }
  printf("%.10f\n", ans / 2.0);
}