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 MAX = 400;
struct hh {
  double x, y;
} a[MAX];
double jisuan(hh a, hh b, hh c) {
  double x1, y1, x2, y2;
  x1 = b.x - a.x;
  y1 = b.y - a.y;
  x2 = c.x - a.x;
  y2 = c.y - a.y;
  return (x1 * y2 - x2 * y1) / 2;
}
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i].x >> a[i].y;
  }
  double area, l, r;
  area = 0;
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      l = r = 0;
      for (int k = 0; k < n; k++) {
        double w = jisuan(a[i], a[j], a[k]);
        if (w > 0 && w > l)
          l = w;
        else if (w < 0 && -w > r)
          r = -w;
      }
      if (l + r > area && l > 0 && r > 0) area = l + r;
    }
  }
  printf("%.6f\n", area);
  return 0;
}