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;
int n, x, y;
pair<int, int> pts[300];
double d[300][300], mx, a1, a2, tmp1, tmp2;
int ccw(pair<int, int> a, pair<int, int> b, pair<int, int> c) {
  int area = (b.first - a.first) * (c.second - a.second) -
             (b.second - a.second) * (c.first - a.first);
  if (area > 0)
    return -1;
  else if (area < 0)
    return 1;
  return 0;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> x >> y;
    pts[i] = make_pair(x, y);
  }
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      double tmp = sqrt(
          ((pts[i].first - pts[j].first) * (pts[i].first - pts[j].first)) +
          ((pts[i].second - pts[j].second) * (pts[i].second - pts[j].second)));
      d[i][j] = d[j][i] = tmp;
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      a1 = a2 = 0;
      for (int k = 0; k < n; k++) {
        if (k != i and k != j) {
          int turn = ccw(pts[k], pts[i], pts[j]);
          if (turn > 0) {
            double s = ((d[i][k] + d[k][j] + d[j][i]) / 2);
            tmp1 = sqrt(s * (s - d[i][k]) * (s - d[k][j]) * (s - d[j][i]));
            if (tmp1 > a1) {
              a1 = tmp1;
            }
          } else if (turn < 0) {
            double s = ((d[i][k] + d[k][j] + d[j][i]) / 2);
            tmp2 = sqrt(s * (s - d[i][k]) * (s - d[k][j]) * (s - d[j][i]));
            if (tmp2 > a2) {
              a2 = tmp2;
            }
          }
        }
      }
      if ((a1 + a2) > mx and a1 > 0 and a2 > 0) {
        mx = a1 + a2;
      }
    }
  }
  printf("%.17lf\n", mx);
  return 0;
}