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 n, ans;
struct Point {
  int x, y;
  Point(int _x = 0, int _y = 0) : x(_x), y(_y) {}
  Point operator-(const Point& _) const { return Point(x - _.x, y - _.y); }
  int operator^(const Point& _) const { return x * _.y - y * _.x; }
};
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  vector<Point> p(n, Point());
  for (int i = 0; i < n; ++i) {
    int x, y;
    cin >> x >> y;
    p[i] = Point(x, y);
  }
  for (int i = 0; i < n; ++i) {
    for (int j = i + 1; j < n; ++j) {
      int mxL = -1, mxR = -1;
      for (int k = 0; k < n; ++k)
        if (i != k && j != k) {
          if (((p[k] - p[i]) ^ (p[j] - p[i])) >= 0)
            mxL = max(mxL, (p[k] - p[i]) ^ (p[j] - p[i]));
          else
            mxR = max(mxR, abs((p[k] - p[i]) ^ (p[j] - p[i])));
        }
      if (mxL != -1 && mxR != -1) ans = max(ans, mxL + mxR);
    }
  }
  cout << fixed << setprecision(10) << (double)ans / 2;
  return 0;
}