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 inf = 1 << 30;
const int maxn = 3333;
const double eps = 1e-10;
int n;
struct point {
  double x, y;
  point() {}
  point(double xx, double yy) : x(xx), y(yy) {}
} P[maxn];
point operator+(point a, point b) { return point(a.x + b.x, a.y + b.y); }
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; }
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%lf%lf", &P[i].x, &P[i].y);
  double ans = 0.0;
  for (int i = 1; i <= n; i++) {
    for (int j = i + 1; j <= n; j++) {
      double lmx = 0, rmx = 0;
      for (int k = 1; k <= n; k++) {
        if (k == i || k == j) continue;
        double cs = cross(P[i] - P[k], P[j] - P[k]) * 0.5;
        if (cs < eps)
          lmx = max(lmx, -cs);
        else
          rmx = max(rmx, cs);
      }
      if (lmx == 0 || rmx == 0) continue;
      ans = max(ans, lmx + rmx);
    }
  }
  printf("%.10f\n", ans);
  return 0;
}