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 main() {
  int n;
  cin >> n;
  double x[n], y[n];
  for (int i = 0; i < n; i++) {
    cin >> x[i] >> y[i];
  }
  double ans = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (i != j) {
        vector<double> dan, bam, zero;
        for (int k = 0; k < n; k++) {
          if (k != i and k != j) {
            double position = ((x[j] - x[i]) * (y[k] - y[i]) -
                               (y[j] - y[i]) * (x[k] - x[i])) *
                              0.5;
            if (position < 0) {
              bam.push_back(abs(position));
            } else if (position > 0) {
              dan.push_back(abs(position));
            } else {
              zero.push_back(position);
            }
          }
        }
        double mxdan = -1, mxbam = -1, z = 0;
        if (dan.size() > 0) {
          for (int i = 0; i < dan.size(); i++) {
            mxdan = max(mxdan, dan[i]);
          }
        }
        if (bam.size() > 0) {
          for (int i = 0; i < bam.size(); i++) {
            mxbam = max(mxbam, bam[i]);
          }
        }
        if (zero.size() > 0) {
          z = 1;
        }
        if (mxdan > 0 and mxbam > 0) {
          ans = max(ans, mxdan + mxbam);
        } else if (z > 0 and (mxbam > -1 || mxdan > -1)) {
          ans = max(ans, max(mxdan, mxbam));
        }
      }
    }
  }
  cout << fixed << setprecision(10) << ans << endl;
}