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 = 1e5 + 10;
int n;
complex<int> s[MAXN];
int cross(complex<int> a, complex<int> b) { return imag(conj(a) * b); }
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  int x, y;
  for (int i = 0; i < n; i++) {
    cin >> x >> y;
    s[i] = complex<int>(x, y);
  }
  double tot = -1;
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      complex<int> a = s[i];
      complex<int> b = s[j];
      double maxi1 = -1;
      for (int k = 0; k < n; k++) {
        complex<int> c = s[k];
        double cc = cross(a - b, c - b);
        if (cc == 0) continue;
        if (cc > 0) maxi1 = max(maxi1, fabs(cc) / 2.0);
      }
      double maxi2 = -1;
      for (int k = 0; k < n; k++) {
        complex<int> c = s[k];
        double cc = cross(a - b, c - b);
        if (cc == 0) continue;
        if (cc < 0) maxi2 = max(maxi2, fabs(cc) / 2.0);
      }
      if (maxi1 != -1 && maxi2 != -1) tot = max(tot, maxi1 + maxi2);
    }
  }
  cout << setprecision(10) << fixed << tot << endl;
  return 0;
}