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>
#pragma GCC optimize("O3")
const double PI = acos(-1.0);
using namespace std;
const int N = 100007, M = 1000000;
const double EPS = 1e-9;
int n, a[N], b[N];
double get(vector<complex<double> > &p) {
  double area = 0;
  int n = p.size();
  for (int i = 0; i < n; i++)
    area += p[i].real() * p[(i + 1) % n].imag() -
            p[i].imag() * p[(i + 1) % n].real();
  return fabs(area) / 2;
}
int ccw(complex<double> a, complex<double> b, complex<double> c) {
  complex<double> v1(b - a), v2(c - a);
  double t = ((conj(v1) * (v2)).imag());
  if (t > +EPS) return +1;
  if (t < -EPS) return -1;
  if (v1.real() * v2.real() < -EPS || v1.imag() * v2.imag() < -EPS) return -1;
  if (norm(v1) < norm(v2) - EPS) return +1;
  return 0;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  ;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i] >> b[i];
  }
  double ans = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = i + 1; j <= n; j++) {
      complex<double> A(a[i], b[i]);
      complex<double> B(a[j], b[j]);
      vector<complex<double> > v;
      v.push_back(A);
      v.push_back(B);
      double dis1 = 0, dis2 = 0;
      for (int k = 1; k <= n; k++) {
        if (k == i || k == j) continue;
        complex<double> C(a[k], b[k]);
        v.push_back(C);
        if (ccw(A, B, C) == 1)
          dis1 = max(dis1, get(v));
        else
          dis2 = max(dis2, get(v));
        v.pop_back();
      }
      if (dis1 > EPS && dis2 > EPS) ans = max(ans, dis1 + dis2);
    }
  }
  cout << fixed << setprecision(9) << ans << endl;
  return 0;
}