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 {
  double x, y;
};
double area(Point a, Point b, Point c) {
  return 0.5 * (a.x * b.y + b.x * c.y + c.x * a.y - a.y * b.x - b.y * c.x -
                c.y * a.x);
}
bool ccw(Point a, Point b, Point c) { return area(a, b, c) > 0; }
int main() {
  int n;
  while (scanf("%d", &n) != EOF) {
    Point arr[n];
    for (int i = 0; i < n; i++) scanf("%lf %lf", &arr[i].x, &arr[i].y);
    double ans = 0;
    for (int i = 0; i < n; i++) {
      for (int j = i + 1; j < n; j++) {
        double x = 0, y = 0;
        for (int k = 0; k < n; k++) {
          if (k == i || k == j) continue;
          double temp = abs(area(arr[i], arr[k], arr[j]));
          if (ccw(arr[i], arr[k], arr[j])) {
            x = max(x, temp);
          } else
            y = max(y, temp);
        }
        if (abs(x) < 1e-10 || abs(y) < 1e-10) continue;
        ans = max(ans, x + y);
      }
    }
    printf("%.6f\n", ans);
  }
}