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;
long long gcd(long long a, long long b) {
  if (a < b) return gcd(b, a);
  if (b == 0) return a;
  return gcd(b, a % b);
}
int square(const vector<pair<int, int> >& r, int a, int b, int c, int d) {
  int sum = r[a].first * r[b].second - r[a].second * r[b].first +
            r[b].first * r[c].second - r[b].second * r[c].first +
            r[c].first * r[d].second - r[c].second * r[d].first +
            r[d].first * r[a].second - r[d].second * r[a].first;
  return max(sum, -sum);
}
int main() {
  int n;
  cin >> n;
  vector<pair<int, int> > r;
  for (int i = 0; i < n; ++i) {
    int x;
    int y;
    cin >> x >> y;
    r.push_back(make_pair(x, y));
  }
  int maximal = 0;
  for (int a = 0; a < n; ++a) {
    for (int c = 0; c < n; ++c) {
      if (a == c) continue;
      int maximalB = -1;
      int maximalSum = 0;
      for (int b = 0; b < n; ++b) {
        if (b == a || b == c) continue;
        int sum = r[a].first * r[b].second - r[a].second * r[b].first +
                  r[b].first * r[c].second - r[b].second * r[c].first;
        if (sum > maximalSum) {
          maximalSum = sum;
          maximalB = b;
        }
      }
      if (maximalB < 0) continue;
      for (int d = 0; d < n; ++d) {
        if (d == a || d == maximalB || d == c) continue;
        maximal = max(maximal, square(r, a, maximalB, c, d));
      }
    }
  }
  cout.precision(10);
  cout << maximal / 2.0 << endl;
}