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 cross(pair<int, int> A, pair<int, int> B) {
  return A.first * 1LL * B.second - A.second * 1LL * B.first;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int N, i, j, k;
  double ans = 0;
  cin >> N;
  vector<pair<int, int> > V(N);
  for (i = 0; i < N; ++i) cin >> V[i].first >> V[i].second;
  for (i = 0; i < N; ++i)
    for (j = i + 1; j < N; ++j) {
      pair<int, int> V1 = {V[i].first - V[j].first, V[i].second - V[j].second};
      double l_1 = 0, l_2 = 0;
      for (k = 0; k < N; ++k)
        if (i != k && j != k &&
            cross(V1, {V[k].first - V[j].first, V[k].second - V[j].second}) > 0)
          l_1 = max(l_1, 0.5 * cross(V1, {V[k].first - V[j].first,
                                          V[k].second - V[j].second}));
        else if (i != k && j != k)
          l_2 = max(l_2, -0.5 * cross(V1, {V[k].first - V[j].first,
                                           V[k].second - V[j].second}));
      if (l_1 > 0 && l_2 > 0) ans = max(ans, l_1 + l_2);
    }
  cout.precision(11);
  cout << fixed << ans << endl;
  return 0;
}