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;
struct TPoint {
  double x, y;
  TPoint(){};
  TPoint(double _x, double _y) {
    x = _x;
    y = _y;
  }
};
struct TVector {
  double x, y;
  TVector() {}
  TVector(TPoint p) {
    x = p.x;
    y = p.y;
  }
  TVector(TPoint p1, TPoint p2) {
    x = p2.x - p1.x;
    y = p2.y - p1.y;
  }
};
double cosv(TVector v1, TVector v2) {
  return (double)(v2.x * v1.y - v1.x * v2.y);
}
int main() {
  int n;
  cin >> n;
  vector<TPoint> in;
  for (int i = 0; i < n; ++i) {
    double x, y;
    cin >> x >> y;
    in.push_back(TPoint(x, y));
  }
  double maxpl = 0;
  for (int i = 0; i < n; ++i) {
    for (int j = i + 1; j < n; ++j) {
      double lefter = -1, righter = -1;
      for (int k = 0; k < n; ++k) {
        if (k != i && k != j) {
          double cosven = cosv(TVector(in[i], in[j]), TVector(in[i], in[k]));
          if (cosven > 0) {
            lefter = max(lefter, cosven / 2);
          } else {
            righter = max(righter, -cosven / 2);
          }
        }
      }
      if (lefter > 0 && righter > 0) maxpl = max(maxpl, lefter + righter);
    }
  }
  cout.precision(10);
  cout << maxpl << endl;
}