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;
pair<int, int> pr[500];
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) scanf("%d%d", &pr[i].first, &pr[i].second);
  int maxi = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (i != j) {
        int max1 = 0, max2 = 0;
        for (int k = 0; k < n; k++) {
          if (k != j && k != i) {
            max1 = min(max1, (pr[i].first * pr[k].second) +
                                 (pr[j].first * pr[i].second) +
                                 (pr[k].first * pr[j].second) -
                                 ((pr[i].first * pr[j].second) +
                                  (pr[j].first * pr[k].second) +
                                  (pr[k].first * pr[i].second)));
            max2 = max(max2, ((pr[i].first * pr[k].second) +
                              (pr[j].first * pr[i].second) +
                              (pr[k].first * pr[j].second)) -
                                 ((pr[i].first * pr[j].second) +
                                  (pr[j].first * pr[k].second) +
                                  (pr[k].first * pr[i].second)));
          }
        }
        if (max1 < 0 && max2 > 0) maxi = max(maxi, -max1 + max2);
      }
    }
  }
  printf("%.12f\n", maxi * 0.5);
  return 0;
}