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;
inline long double ccw(pair<long double, long double> A,
                       pair<long double, long double> B,
                       pair<long double, long double> C) {
  return ((B.first - A.first) * (C.second - B.second) -
          (B.second - A.second) * (C.first - B.first)) /
         2;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long T = 1;
  while (T--) {
    long long n, i, j, k;
    cin >> n;
    vector<pair<long double, long double>> v(n);
    for (auto &x : v) {
      cin >> x.first >> x.second;
    }
    long double ans = 0;
    for (i = 0; i < n; i++) {
      for (j = i + 1; j < n; j++) {
        long double mx = -1;
        long double mn = 1;
        for (k = 0; k < n; k++) {
          if (k == i || k == j) continue;
          long double val = ccw(v[i], v[j], v[k]);
          if (val > 0)
            mx = max(mx, val);
          else
            mn = min(mn, val);
        }
        if (mx > 0 && mn < 0) {
          ans = max(ans, mx - mn);
        }
      }
    }
    cout.precision(9);
    cout << ans << "\n";
  }
  return 0;
}