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;
typedef struct {
  double x, y;
} Point;
typedef struct {
  Point x, y;
} Vector;
Point a[1005];
double xmulti(const Point p1, const Point p2, const Point p0) {
  return (p1.x - p0.x) * (p2.y - p0.y) - (p1.y - p0.y) * (p2.x - p0.x);
}
bool side(const Point p1, const Vector vec) {
  return xmulti(vec.x, p1, vec.y) > 1e-6;
}
int main() {
  int i, j, n, x, y, k;
  double ans = 0;
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%lf%lf", &a[i].x, &a[i].y);
  }
  for (i = 0; i < n; i++) {
    for (j = i + 1; j < n; j++) {
      Vector vec;
      vec.x = a[i];
      vec.y = a[j];
      double s1 = 0, s2 = 0;
      for (k = 0; k < n; k++) {
        if (side(a[k], vec)) {
          s1 = max(s1, fabs(xmulti(a[i], a[j], a[k])) / 2);
        } else {
          s2 = max(s2, fabs(xmulti(a[i], a[j], a[k])) / 2);
        }
      }
      if (s1 != 0 && s2 != 0) ans = max(ans, s1 + s2);
    }
  }
  printf("%lf\n", ans);
  return 0;
}