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>
#pragma comment(linker, "/STACK:100000000000,100000000000")
using namespace std;
const long long inf = 1e18 + 7;
const long long mod = 1e9 + 7;
const double eps = 1e-12;
const double PI = 2 * acos(0.0);
const double E = 2.71828;
long long n;
struct pt {
  long long x, y;
} a[100005];
long long second(pt a, pt b, pt c) {
  return (b.x - a.x) * (c.y - a.y) - (c.x - a.x) * (b.y - a.y);
}
long long res = 0;
int main(void) {
  cin >> n;
  for (int(i) = 0; (i) < (long long)(n); (i)++) cin >> a[i].x >> a[i].y;
  for (int(i) = 0; (i) < (long long)(n); (i)++)
    for (int(j) = 0; (j) < (long long)(i); (j)++) {
      long long l = 0, r = 0;
      for (int(k) = 0; (k) < (long long)(n); (k)++)
        if (k != i && k != j) {
          long long ns = second(a[i], a[j], a[k]);
          if (ns < 0)
            l = max(l, -ns);
          else
            r = max(r, ns);
        }
      if (l && r) res = max(res, l + r);
    }
  cout << setprecision(12) << fixed << res / 2.0 << endl;
  return 0;
}