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 operator-(point &other) { return point(x - other.x, y - other.y); }
};
const int N = 305;
point arr[N];
long long cross(point v1, point v2) {
  return 1ll * v1.x * v2.y - 1ll * v1.y * v2.x;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d %d", &arr[i].x, &arr[i].y);
  }
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      long long Max = -1e18, Min = 1e18;
      for (int k = 0; k < n; k++) {
        if (i == j || j == k || k == i) continue;
        point &a = arr[i];
        point &b = arr[j];
        point &c = arr[k];
        long long tarea = cross(b - a, c - a);
        if (tarea < Min) {
          Min = tarea;
        }
        if (tarea > Max) {
          Max = tarea;
        }
        if (abs(Max - Min) > ans) {
          ans = abs(Max - Min);
        }
      }
    }
  }
  printf("%.8f\n", 1.0 * ans / 2.0);
}