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 pix {
  int x, y;
  pix(int x, int y) : x(x), y(y){};
  int operator*(pix a) { return a.y * x - a.x * y; }
  pix operator-(pix a) { return pix(x - a.x, y - a.y); }
};
bool operator<(pix a, pix b) {
  if (a.x * b.y - a.y * b.x < 0) return true;
  if (a.x * b.y == a.y * b.x &&
      a.x * a.x + a.y * a.y - b.x * b.x - b.y * b.y < 0)
    return true;
  return false;
}
int main() {
  int k;
  vector<pix> a;
  cin >> k;
  for (int i = 0; i < k; i++) {
    int x, y;
    cin >> x >> y;
    a.push_back(pix(x, y));
  }
  int s = 0;
  int up = 0, down = 0;
  for (int i = 0; i < a.size(); i++)
    for (int j = 0; j < a.size(); j++)
      if (i != j) {
        for (int k = 0; k < a.size(); k++) {
          if (up < (a[i] - a[j]) * (a[k] - a[j]))
            up = (a[i] - a[j]) * (a[k] - a[j]);
          if (down > (a[i] - a[j]) * (a[k] - a[j]))
            down = (a[i] - a[j]) * (a[k] - a[j]);
        }
        if (down && up) s = max(s, up - down);
        up = down = 0;
      }
  cout.precision(10);
  cout << fixed << s * 1.0 / 2;
}