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 maxn = 300;
pair<long long int, long long int> p[maxn];
long long int cross(pair<long long int, long long int> A,
                    pair<long long int, long long int> B) {
  return A.first * B.second - A.second * B.first;
}
pair<long long int, long long int> mnus(pair<long long int, long long int> A,
                                        pair<long long int, long long int> B) {
  return {A.first - B.first, A.second - B.second};
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> p[i].first >> p[i].second;
  long long int ans = 0;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < i; j++) {
      pair<long long int, long long int> A = p[i], B = p[j];
      long long int max_neg = -1, max_pos = -1;
      for (int k = 0; k < n; k++)
        if (k != i && k != j) {
          long long int c = cross(mnus(p[k], A), mnus(B, A));
          if (c < 0) max_neg = max(max_neg, abs(c));
          if (c > 0) max_pos = max(max_pos, abs(c));
        }
      if (max_pos != -1 && max_neg != -1) ans = max(ans, max_neg + max_pos);
    }
  cout.precision(12);
  cout << fixed << (long double)ans / (long double)2 << endl;
  return 0;
}