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 dx[] = {0, 1, -1, 0};
int dy[] = {1, 0, 0, -1};
long long pw(long long b, long long p) {
  if (!p) return 1;
  long long sq = pw(b, p / 2);
  sq *= sq;
  if (p % 2) sq *= b;
  return sq;
}
const int N = 302, M = (N << 1);
int x[N], y[N];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> x[i] >> y[i];
  }
  double res = 0;
  for (int i = 0; i < n - 1; i++) {
    for (int j = i + 1; j < n; j++) {
      double mxl = 0, mxr = 0;
      for (int k = 0; k < n; k++) {
        if (i == k || j == k) continue;
        double area =
            ((x[i] - x[k]) * (y[j] - y[k]) - (y[i] - y[k]) * (x[j] - x[k])) /
            2.0;
        if (area > 0) mxl = max(mxl, area);
        if (area < 0) mxr = max(mxr, -area);
        if (mxl != 0 && mxr != 0) {
          res = max(res, mxl + mxr);
        }
      }
    }
  }
  cout << fixed << setprecision(6) << res;
  return 0;
}