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 inf = 0x3f3f3f3f;
const long long infll = 0x3f3f3f3f3f3f3f3f;
const long long mod = 1e9 + 7;
const double eps = 1e-8;
const double pi = acos(-1.0);
const int maxn = 1e5 + 5;
const int maxm = 1.5e7 + 5;
using namespace std;
struct node {
  double x, y;
} p[305];
int n;
double cal(node p0, node p1, node p2) {
  return (p0.x - p1.x) * (p2.y - p1.y) - (p0.y - p1.y) * (p2.x - p1.x);
}
int main() {
  double ans = 0.0;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> p[i].x >> p[i].y;
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      double ans1 = 0, ans2 = 0;
      for (int k = 1; k <= n; k++) {
        double tmp = cal(p[i], p[j], p[k]);
        if (tmp > 0)
          ans1 = max(ans1, tmp / 2.0);
        else if (tmp < 0)
          ans2 = max(ans2, -tmp / 2.0);
      }
      if (ans1 > 0 && ans2 > 0) ans = max(ans, ans1 + ans2);
    }
  }
  printf("%.8f\n", ans);
  return 0;
}