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 maxn = 100000 + 50;
const double eps = 1e-10;
int gcd(int a, int b) { return a % b == 0 ? b : gcd(b, a % b); }
struct Point {
  double x, y;
  Point() {}
  Point(double x, double y) : x(x), y(y) {}
  Point operator-(Point B) { return Point(x - B.x, y - B.y); }
} p[500];
double Cross(Point A, Point B) { return A.x * B.y - A.y * B.x; }
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%lf %lf", &p[i].x, &p[i].y);
  double ans = 0;
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      double left = 0, right = 0;
      for (int k = 0; k < n; k++) {
        if (k == i || k == j) continue;
        double s = Cross(p[i] - p[k], p[j] - p[k]) * 0.5;
        if (s < eps)
          left = max(left, -s);
        else
          right = max(right, s);
      }
      if (left == 0 || right == 0) continue;
      ans = max(left + right, ans);
    }
  }
  printf("%lf\n", ans);
  return 0;
}