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;
double ccw(pair<int, int> A, pair<int, int> B, pair<int, int> C) {
  double ans = 0.0;
  double x1, x2, y1, y2;
  x1 = B.first - A.first;
  y1 = B.second - A.second;
  x2 = C.first - A.first;
  y2 = C.second - A.second;
  ans = ((double)x1 * y2 - (double)x2 * y1) * .5;
  return ans;
}
double eps = 0.00000001;
double cmax(double A, double B) {
  if (B - A > eps)
    return B;
  else
    return A;
}
int main() {
  int n;
  cin >> n;
  vector<pair<int, int> > A(n);
  for (int i = 0; i < (n); ++i) cin >> A[i].first >> A[i].second;
  double Ans = 0.0;
  for (int i = 0; i < (n); ++i) {
    for (int j = i + 1; j < (n); ++j) {
      double ma = -1, mb = -1;
      for (int k = 0; k < (n); ++k) {
        if (i == k || j == k) continue;
        double tmp = ccw(A[i], A[j], A[k]);
        if (tmp < 0)
          ma = cmax(ma, -tmp);
        else
          mb = cmax(mb, tmp);
      }
      if (ma != -1 && mb != -1) Ans = cmax(Ans, ma + mb);
    }
  }
  printf("%.7lf\n", Ans);
  return 0;
}