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;
  Point(double x = 0, double y = 0) : x(x), y(y) {}
  Point operator-(const Point &b) const { return Point(x - b.x, y - b.y); }
  Point operator+(const Point &b) const { return Point(x + b.x, y + b.y); }
  double operator*(const Point &b) const { return x * b.y - y * b.x; }
  double len() { return sqrt(x * x + y * y); }
} A[333];
int n;
double ans[2];
int main() {
  cin >> n;
  for (int i = 1; i <= n; ++i) scanf("%lf%lf", &A[i].x, &A[i].y);
  double res = 0;
  for (int i = 1; i <= n; ++i)
    for (int j = i + 1; j <= n; ++j) {
      ans[0] = ans[1] = -1;
      for (int k = 1; k <= n; ++k)
        if (k != i && k != j) {
          double &t = ans[(A[k] - A[i]) * (A[j] - A[i]) < 0];
          t = max(t, fabs((A[k] - A[i]) * (A[j] - A[i])) / 2.0);
        }
      if (ans[0] >= 0 && ans[1] >= 0) res = max(res, ans[0] + ans[1]);
    }
  printf("%.12f\n", res);
  return 0;
}