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 nMax = 305;
int n;
struct Point {
  int x, y;
} v[nMax];
inline long double cross_product(const Point &o, const Point &a,
                                 const Point &b) {
  return (a.x - o.x) * (b.y - o.y) - (a.y - o.y) * (b.x - o.x);
}
inline long double Surface(const Point &a, const Point &b, const Point &c) {
  return 0.5 * cross_product(a, b, c);
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> v[i].x >> v[i].y;
  long double mx1, mx2, t, ans = 0;
  for (int i = 1; i <= n; ++i)
    for (int j = i + 1; j <= n; ++j) {
      mx1 = mx2 = 0;
      for (int k = 1; k <= n; ++k) {
        if (k == i || k == j) continue;
        t = Surface(v[i], v[j], v[k]);
        if (t < 0)
          mx1 = max(mx1, -t);
        else
          mx2 = max(mx2, t);
      }
      if (mx1 == 0 || mx2 == 0) continue;
      ans = max(ans, mx1 + mx2);
    }
  cout << setprecision(12) << fixed << ans;
  return 0;
}