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;
int n;
pair<int, int> p[305];
int func(int x1, int y1, int x2, int y2, int x3, int y3) {
  return (x2 - x1) * (y3 - y1) - (y2 - y1) * (x3 - x1);
}
double area(int i, int j, int k) {
  double ret = func(p[i].first, p[i].second, p[j].first, p[j].second,
                    p[k].first, p[k].second);
  return ret / 2.0;
}
void oku() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d%d", &p[i].first, &p[i].second);
  sort(p, p + n);
  double ret = 0;
  for (int i = 0; i < n; i++)
    for (int j = i + 1; j < n; j++) {
      double posMx = 0, posMn = 1000000000;
      double negMx = 0, negMn = 1000000000;
      for (int k = 0; k < n; k++)
        if (k != i && k != j) {
          double r = area(i, j, k);
          if (r > 0) {
            posMx = max(posMx, r);
            posMn = min(posMn, r);
          } else {
            negMx = max(negMx, -r);
            negMn = min(negMn, -r);
          }
        }
      ret = max(ret, max(posMx - posMn, negMx - negMn));
      if (negMx > 0 && posMx > 0) ret = max(ret, posMx + negMx);
    }
  printf("%.10f\n", ret);
}
int main() {
  oku();
  return 0;
}