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 N = 500;
int n;
pair<double, double> a[N];
vector<pair<double, double> > l, r;
double area(pair<double, double> a, pair<double, double> b,
            pair<double, double> c) {
  return fabs((b.first - a.first) * (c.second - a.second) -
              (b.second - a.second) * (c.first - a.first)) /
         2.0;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i].first >> a[i].second;
  }
  double s = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      double A = a[j].second - a[i].second;
      double B = a[i].first - a[j].first;
      double C = a[i].first * (-A) + a[i].second * (-B);
      l.clear();
      r.clear();
      for (int k = 1; k <= n; k++) {
        if (A * a[k].first + B * a[k].second + C < 0) l.push_back(a[k]);
        if (A * a[k].first + B * a[k].second + C > 0) r.push_back(a[k]);
      }
      if (!l.size() || !r.size()) continue;
      double s1 = 0;
      for (int k = 0; k < l.size(); k++) {
        s1 = max(s1, area(a[i], a[j], l[k]));
      }
      double s2 = 0;
      for (int k = 0; k < r.size(); k++) {
        s2 = max(s2, area(a[i], a[j], r[k]));
      }
      s = max(s, s1 + s2);
    }
  }
  printf("%.15f", s);
  return 0;
}