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;
constexpr int N = 1e6 + 10;
struct Point {
  int x, y;
} point[N];
int n;
double ans;
int ccw(int a, int b, int c) {
  return (point[b].x - point[a].x) * (point[c].y - point[a].y) -
         (point[b].y - point[a].y) * (point[c].x - point[a].x);
}
double calc(int a, int b, int c) {
  double k = abs(point[a].x * point[b].y + point[a].y * point[c].x +
                 point[b].x * point[c].y - point[c].x * point[b].y -
                 point[c].y * point[a].x - point[b].x * point[a].y);
  k /= 2;
  return k;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> point[i].x >> point[i].y;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (i == j) continue;
      double chap = 0, rast = 0;
      for (int k = 0; k < n; k++) {
        if (k == i || k == j) continue;
        if (ccw(i, j, k)) {
          if (ccw(i, j, k) > 0)
            rast = max(rast, calc(i, j, k));
          else
            chap = max(chap, calc(i, j, k));
        }
      }
      if (chap && rast) ans = max(ans, rast + chap);
    }
  }
  printf("%.8f\n", ans);
  return 0;
}