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 double INF = 1000000000;
struct node {
  double x, y;
};
node p[350];
double mul(int i, int j, int k) {
  return ((p[k].x - p[i].x) * (p[k].y - p[j].y) -
          (p[k].y - p[i].y) * (p[k].x - p[j].x)) /
         2.0;
}
int main() {
  int n, i, j, k;
  while (scanf("%d", &n) != EOF) {
    for (i = 0; i < n; i++) {
      scanf("%lf%lf", &p[i].x, &p[i].y);
    }
    double lmax = -INF, rmax = -INF, amax = -INF;
    for (i = 0; i < n; i++) {
      for (j = 0; j < n; j++) {
        if (i == j) continue;
        lmax = -INF, rmax = -INF;
        for (k = 0; k < n; k++) {
          if (i == k || j == k) continue;
          double temp = mul(i, j, k);
          if (temp < 0) {
            lmax = max(lmax, -temp);
          } else {
            rmax = max(rmax, temp);
          }
        }
        amax = max(amax, lmax + rmax);
      }
    }
    printf("%.6f\n", amax);
  }
  return 0;
}