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 ww {
  double a, b;
} a[310];
int i, j, k, n, m;
double an;
inline double cross(int i, int j, int k) {
  return (a[j].a - a[i].a) * (a[k].b - a[i].b) -
         (a[j].b - a[i].b) * (a[k].a - a[i].a);
}
int main() {
  scanf("%d", &n);
  for (i = 1; i <= n; i++) scanf("%lf%lf", &a[i].a, &a[i].b);
  for (i = 1; i <= n; i++)
    for (j = i + 1; j <= n; j++) {
      double mi, ma;
      mi = ma = 0;
      for (k = 1; k <= n; k++) {
        double A = cross(i, j, k);
        mi = min(mi, A), ma = max(ma, A);
      }
      if (!mi) mi = 1e15;
      if (!ma) ma = -1e15;
      an = max(an, ma - mi);
    }
  printf("%.12lf\n", an / 2);
  return 0;
}