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>
const int MAX_N = 10001;
using namespace std;
const int mod = 1000000007;
const int MAXN = 333;
const double eps = 1e-10;
struct Point {
  double x, y;
  Point(double _x = 0, double _y = 0) : x(_x), y(_y) {}
} p[MAXN];
Point operator-(Point a, Point b) { return Point(a.x - b.x, a.y - b.y); }
double cross(Point a, Point b) { return (a.x * b.y - a.y * b.x) * 0.5; }
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%lf%lf", &p[i].x, &p[i].y);
  }
  double ans = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = i + 1; j <= n; j++) {
      double lm, rm;
      lm = rm = 0;
      for (int k = 1; k <= n; k++) {
        if (k == i || k == j) continue;
        double s = cross(p[i] - p[k], p[j] - p[k]);
        if (s < eps)
          lm = max(lm, -s);
        else
          rm = max(rm, s);
      }
      if (lm == 0 || rm == 0) continue;
      ans = max(ans, lm + rm);
    }
  }
  printf("%f\n", ans);
  return 0;
}