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 {
  int x, y;
  point(int x = 0, int y = 0) : x(x), y(y) {}
};
point p[305];
int n;
point operator-(point a, point b) { return point(a.x - b.x, a.y - b.y); }
int cp(point a, point b) { return a.x * b.y - a.y * b.x; }
double work() {
  int i, j, k, f1, f2;
  double s1, s2, ans = 0;
  for (i = 1; i <= n; ++i)
    for (j = i + 1; j <= n; ++j) {
      s1 = s2 = 0;
      f1 = f2 = 0;
      for (k = 1; k <= n; ++k)
        if (k != i && k != j) {
          if (cp(p[i] - p[j], p[k] - p[j]) < 0) {
            s1 = max(s1, -(double)cp(p[i] - p[j], p[k] - p[j]) / 2);
            f1 = 1;
          } else {
            s2 = max(s2, (double)cp(p[i] - p[j], p[k] - p[j]) / 2);
            f2 = 1;
          }
        }
      if (f1 && f2) ans = max(ans, s1 + s2);
    }
  return ans;
}
int main() {
  int i;
  scanf("%d", &n);
  for (i = 1; i <= n; ++i) scanf("%d%d", &p[i].x, &p[i].y);
  printf("%.10lf\n", work());
  return 0;
}