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 area(pair<double, double> p1, pair<double, double> p2,
            pair<double, double> p3) {
  return 0.5 * (((p2.first - p1.first) * (p3.second - p2.second)) -
                ((p2.second - p1.second) * (p3.first - p2.first)));
}
int main() {
  int n;
  cin >> n;
  vector<pair<double, double> > points;
  for (int i = 0; i < n; i++) {
    double a, b;
    cin >> a >> b;
    points.push_back(make_pair(a, b));
  }
  double ans = INT_MIN;
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      double mn = -1;
      double mx = -1;
      for (int k = 0; k < n; k++) {
        if (k != i && k != j) {
          double a = area(points[i], points[j], points[k]);
          if (a < 0) {
            mn = max(mn, -a);
          } else {
            mx = max(mx, a);
          }
        }
      }
      if (mn >= 0 && mx >= 0) {
        ans = max(ans, mn + mx);
      }
    }
  }
  printf("%lf", ans);
  return 0;
}