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;
template <typename T>
inline T sqr(T n) {
  return (n * n);
}
int n;
pair<int, int> a[1000];
bool lo(pair<int, int> t, pair<int, int> a1, pair<int, int> b) {
  pair<int, int> aa(make_pair(a1.first - t.first, a1.second - t.second));
  pair<int, int> bb(make_pair(b.first - t.first, b.second - t.second));
  return (aa.first * bb.second - aa.second * bb.first < 0);
}
double second(int a1, int b1, int c1) {
  pair<int, int> aa =
      make_pair(a[b1].first - a[a1].first, a[b1].second - a[a1].second);
  pair<int, int> bb =
      make_pair(a[c1].first - a[a1].first, a[c1].second - a[a1].second);
  return abs(aa.first * bb.second - aa.second * bb.first);
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d%d", &a[i].first, &a[i].second);
  double res = 0;
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      double mx = -1, mx1 = -1;
      for (int k = 0; k < n; k++) {
        if (i != j && i != k && j != k) {
          if (lo(a[i], a[j], a[k]))
            mx = max(mx, (second(i, j, k)));
          else
            mx1 = max(mx1, (second(i, j, k)));
        }
      }
      if (mx != -1 && mx1 != -1) res = max(res, mx + mx1);
    }
  }
  printf("%.14lf", res / 2);
  return 0;
}