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 maxn = 3e2 + 12;
int x[maxn], y[maxn], n;
double save[maxn][maxn];
int cross(int a, int b, int c) {
  return (x[b] - x[a]) * (y[c] - y[a]) - (y[b] - y[a]) * (x[c] - x[a]);
}
double dis(int a, int b) {
  if (save[a][b]) return save[a][b];
  return save[a][b] = sqrt((y[a] - y[b]) * (y[a] - y[b]) +
                           (x[a] - x[b]) * (x[a] - x[b]));
}
double s(int a, int b, int c) {
  double p = (dis(a, b) + dis(b, c) + dis(c, a)) / 2.0;
  return sqrt(p * (p - dis(a, b)) * (p - dis(a, c)) * (p - dis(c, b)));
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d%d", x + i, y + i);
  double l = 0, r = 0, ans = 0;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n;
         j++, l && r ? (ans = max(ans, l + r), l = r = 0) : l = r = 0)
      if (i != j)
        for (int k = 0; k < n; k++)
          if (i != k && j != k)
            cross(i, j, k) > 0 ? l = max(l, s(i, j, k))
                               : r = max(r, s(i, j, k));
  cout << fixed << setprecision(9) << ans << '\n';
  return 0;
}