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;
const long double eps = 0.00000001;
struct point {
  long double x, y;
  point() {}
  point(long double _x, long double _y) {
    x = _x;
    y = _y;
  }
};
point mas[500];
long double antipsevdo(point a, point b, point c) {
  return ((b.x - a.x) * (c.y - b.y) - (b.y - a.y) * (c.x - b.x)) * 0.5;
}
void maximum(long double &a, long double b) {
  if (b - a > eps) a = b;
  return;
}
int main() {
  int n;
  cin >> n;
  long double maxZ = 0;
  for (int i = 0; i < n; i++) {
    cin >> mas[i].x >> mas[i].y;
  }
  for (int i = 0; i < n; i++)
    for (int j = i + 1; j < n; j++) {
      long double minn = -1, maxx = -1;
      for (int k = 0; k < n; k++)
        if (k != i && k != j)
          if (antipsevdo(mas[i], mas[j], mas[k]) < 0)
            maximum(minn, -antipsevdo(mas[i], mas[j], mas[k]));
          else
            maximum(maxx, antipsevdo(mas[i], mas[j], mas[k]));
      if (maxx >= 0 && minn >= 0) maximum(maxZ, maxx + minn);
    }
  cout.precision(20);
  cout << maxZ;
  return 0;
}