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 double cross(const pair<int, int> &O, const pair<int, int> &A,
                    const pair<int, int> &B) {
  return 0.5 * ((A.first - O.first) * (double)(B.second - O.second) -
                (A.second - O.second) * (double)(B.first - O.first));
}
inline double area(vector<pair<int, int> > p, pair<int, int> a) {
  double res = 0.0;
  p.push_back(a);
  p.push_back(p[0]);
  for (int i = int(0); i < int(p.size() - 1); ++i) {
    res += (p[i].first * p[i + 1].second - p[i + 1].first * p[i].second);
  }
  return fabs(res) / 2.0;
}
int main() {
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  vector<pair<int, int> > v;
  for (int i = int(0); i < int(n); ++i) {
    pair<int, int> a;
    cin >> a.first >> a.second;
    v.push_back(a);
  }
  double ans = 0;
  for (int i = int(0); i < int(n); ++i) {
    for (int j = int(i + 1); j < int(n); ++j) {
      double t1 = 0, t2 = 0;
      for (int k = int(0); k < int(n); ++k) {
        if (k == i || k == j) continue;
        double d = cross(v[i], v[j], v[k]);
        if (d < 0) {
          t1 = max(t1, -d);
        } else {
          t2 = max(t2, d);
        }
      }
      if (t1 < 0.000001 || t2 < 0.000001) continue;
      ans = max(ans, t1 + t2);
    }
  }
  cout << fixed << setprecision(12) << ans << endl;
  return 0;
}