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;
int x[2010], y[2010];
int n;
double find(int i, int j) {
  int x1 = x[i], y1 = y[i], x2 = x[j], y2 = y[j];
  int a = x1 - x2, b = y1 - y2;
  int c = -b * x1 + a * y1;
  int maxp = 0, maxn = 0, t;
  for (int k = 0; k < n; k++) {
    t = b * x[k] - a * y[k] + c;
    if (t > 0 && t > maxp) maxp = t;
    if (t < 0 && -t > maxn) maxn = -t;
  }
  if (maxp > 0 && maxn > 0)
    return ((maxp + maxn) / 2.0);
  else
    return 0;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> x[i] >> y[i];
  }
  double res = 0, tem;
  for (int i = 0; i < n - 1; i++)
    for (int j = i + 1; j < n; j++) {
      tem = find(i, j);
      if (tem > res) res = tem;
    }
  printf("%.6f", res);
  return 0;
}