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 a[304];
int fre[304];
vector<pair<double, double> > v;
double ab(double x) {
  if (x < 0) x *= (double)-1;
  return x;
}
double val(double x, double y, double z) { return x * (y - z); }
int main(int argc, char const *argv[]) {
  double ans = 0;
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    double a, b;
    cin >> a >> b;
    v.push_back(make_pair(a, b));
  }
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      double d1 = v[j].first - v[i].first;
      double d2 = v[j].second - v[i].second;
      double m = d2 / d1;
      double c = v[i].second - m * v[i].first;
      double mx = 0;
      int p = -1, q = -1;
      for (int k = 0; k < n; k++) {
        if (k != i && k != j && v[k].second - m * v[k].first - c > mx) {
          mx = v[k].second - m * v[k].first - c;
          p = k;
        }
      }
      mx = 0;
      for (int k = 0; k < n; k++) {
        if (k != i && k != j && k != p &&
            v[k].second - m * v[k].first - c < mx) {
          mx = v[k].second - m * v[k].first - c;
          q = k;
        }
      }
      if (p != -1 && q != -1) {
        double a1 = ab(val(v[i].first, v[j].second, v[p].second) +
                       val(v[j].first, v[p].second, v[i].second) +
                       val(v[p].first, v[i].second, v[j].second));
        a1 /= 2.0;
        double a2 = ab(val(v[i].first, v[j].second, v[q].second) +
                       val(v[j].first, v[q].second, v[i].second) +
                       val(v[q].first, v[i].second, v[j].second));
        a2 /= 2.0;
        ans = max(ans, a1 + a2);
      }
    }
  }
  printf("%.10lf", ans);
  return 0;
}