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 point {
  long long x, y;
};
point p[300];
long long inf = (1LL << 60);
int ccw(point& a, point& b, point& c) {
  return (a.x * b.y + b.x * c.y + c.x * a.y) -
         (a.y * b.x + b.y * c.x + c.y * a.x);
}
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> p[i].x >> p[i].y;
  }
  long long sol = 0;
  for (int i = 0; i < n; ++i) {
    for (int j = i + 1; j < n; ++j) {
      long long tmp;
      long long minv = inf;
      long long maxv = -inf;
      for (int k = 0; k < n; ++k) {
        if (k == i || k == j) {
          continue;
        }
        tmp = ccw(p[k], p[i], p[j]);
        maxv = max(maxv, tmp);
        minv = min(minv, tmp);
      }
      if (maxv <= 0 || minv >= 0) {
        continue;
      }
      sol = max(sol, maxv - minv);
    }
  }
  double s = sol / 2.0;
  printf("%lf\n", s);
  return 0;
}