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;
const int N = 305;
int n, x[N], y[N], a[N], ans;
int f(int i, int j, int k, int l) {
  return x[i] * y[j] + x[j] * y[k] + x[k] * y[l] + x[l] * y[i] - x[j] * y[i] -
         x[k] * y[j] - x[l] * y[k] - x[i] * y[l];
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> x[i] >> y[i];
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if (i == j) continue;
      int dx = x[j] - x[i];
      int dy = y[i] - y[j];
      pair<int, int> now = {-1, -1};
      for (int k = 1; k <= n; k++) {
        if (k == i || k == j) continue;
        a[k] = x[k] * dy + y[k] * dx;
        if (now.first == -1 || a[k] >= a[now.first]) {
          now.second = now.first;
          now.first = k;
        } else if (now.second == -1 || a[k] >= a[now.second]) {
          now.second = k;
        }
      }
      for (int k = 1; k <= n; k++) {
        if (k == i || k == j) continue;
        int l = now.first;
        if (l == k) {
          l = now.second;
        }
        ans = max(ans, f(i, k, j, l));
      }
    }
  }
  cout << ans / 2 << "." << ((ans & 1) ? 5 : 0) << "\n";
}