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 int M = 1e5 + 10;
const int Q = 1e9 + 7;
const int N = 1e7;
const long double eps = 1e-9;
int x[500], y[500];
long double dist(int i, int j) {
  return sqrt((x[i] - x[j]) * (x[i] - x[j]) + (y[i] - y[j]) * (y[i] - y[j]) +
              0.);
}
int main() {
  srand(time(NULL));
  int n, i, j, t;
  long long a, b, c, ma1, ma2, cur;
  long double ans = 0.;
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%d%d", &x[i], &y[i]);
  }
  for (i = 0; i < n; i++)
    for (j = i + 1; j < n; j++) {
      a = y[i] - y[j];
      b = x[j] - x[i];
      c = 0 - a * x[i] - b * y[i];
      ma1 = 0;
      ma2 = 0;
      for (t = 0; t < n; t++) {
        if (t == i || t == j) continue;
        cur =
            (long long)a * (long long)x[t] + (long long)b * (long long)y[t] + c;
        if (cur > 0)
          ma1 = max(ma1, cur);
        else
          ma2 = min(ma2, cur);
      }
      if (ma1 == 0 || ma2 == 0) continue;
      long double bob = 2 * sqrt((long double)a * a + b * b + 0.);
      ans = max(ans, dist(i, j) * (ma1 - ma2) / bob);
    }
  printf("%.15f", (double)ans);
  return 0;
}