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;
struct pt {
  long long x, y;
};
long long area2(pt a, pt b, pt c) {
  return fabs(a.x * b.y - a.y * b.x + b.x * c.y - b.y * c.x + c.x * a.y -
              c.y * a.x);
}
int turn(pt p, pt q, pt r) {
  long long result = (r.x - q.x) * (p.y - q.y) - (r.y - q.y) * (p.x - q.x);
  if (result < 0) return -1;
  if (result > 0) return 1;
  return 0;
}
bool ccw(pt p, pt q, pt r) { return (turn(p, q, r) > 0); }
int main() {
  ios_base::sync_with_stdio(0);
  ;
  int precision = numeric_limits<double>::digits10;
  ;
  int n;
  cin >> n;
  vector<pt> p(n);
  for (int i = 0; i < n; i++) {
    cin >> p[i].x >> p[i].y;
  }
  long long ans1 = 0;
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      long long cl = -1, acl = -1;
      for (int k = 0; k < n; k++) {
        if (k != i and k != j) {
          if (ccw(p[i], p[j], p[k]))
            cl = max(cl, area2(p[i], p[j], p[k]));
          else
            acl = max(acl, area2(p[i], p[j], p[k]));
        }
      }
      if (cl != -1 and acl != -1) ans1 = max(ans1, cl + acl);
    }
  }
  if (ans1 % 2 == 0)
    cout << ans1 / 2 << endl;
  else
    cout << ans1 / 2 << ".5" << endl;
}