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 = 300;
struct point {
  int x, y;
  point() {}
  point(int x, int y) : x(x), y(y) {}
  point operator-(point o) { return point(x - o.x, y - o.y); }
  double operator*(point o) { return x * o.y - o.x * y; }
} a[MAXN];
int square(point a, point b, point c) { return (c - a) * (c - b); }
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d%d", &a[i].x, &a[i].y);
  }
  int ans = 0;
  for (int fst = 0; fst < n; fst++) {
    for (int snd = 0; snd < n; snd++) {
      if (snd == fst) continue;
      point p(a[fst]);
      point q(a[snd]);
      int mn = INT_MAX;
      int mx = -INT_MAX;
      for (int thd = 0; thd < n; thd++) {
        if (thd == fst || thd == snd) continue;
        point m(a[thd]);
        mn = min(mn, square(p, q, m));
        mx = max(mx, square(p, q, m));
      }
      ans = max(ans, mx - mn);
    }
  }
  printf("%.1lf", ans / 2.0);
  return 0;
}