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>
#pragma GCC optimize("O3")
using namespace std;
const int N = 25 + 5, M = 1e3 + 5, OO = 0x3f3f3f3f;
vector<complex<int> > ps;
int n;
double ans, ar1, ar2;
double calc(const complex<int>& a, const complex<int>& b,
            const complex<int>& c) {
  return ((b.real() - a.real()) * (c.imag() - b.imag()) -
          (b.imag() - a.imag()) * (c.real() - b.real())) *
         0.5;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0, a, b; i < n; i++) cin >> a >> b, ps.emplace_back(a, b);
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      ar1 = ar2 = -OO;
      for (int k = 0; k < n; k++) {
        if (k == i || k == j) continue;
        if (calc(ps[i], ps[j], ps[k]) < 0.0)
          ar1 = max(ar1, -calc(ps[i], ps[j], ps[k]));
        else
          ar2 = max(ar2, calc(ps[i], ps[j], ps[k]));
      }
      ans = max(ans, ar1 + ar2);
    }
  }
  cout << fixed << setprecision(10) << ans;
  return 0;
}