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;
template <class T>
inline bool chmax(T &a, T b) {
  return a < b ? (a = b, 1) : 0;
}
template <class T>
inline bool chmin(T &a, T b) {
  return a > b ? (a = b, 1) : 0;
}
template <class T>
inline void chmaxmin(T &ma, T &mi, T x) {
  chmax(ma, x), chmin(mi, x);
}
using Real = double;
using Point = complex<Real>;
ostream &operator<<(ostream &os, Point &p) {
  return os << real(p) << " " << imag(p);
}
istream &operator>>(istream &is, Point &p) {
  Real x, y;
  is >> x >> y;
  p = Point(x, y);
  return is;
}
Real cross(const Point &a, const Point &b) { return imag(conj(a) * b); }
int main() {
  cout << fixed << setprecision(12);
  int n;
  cin >> n;
  vector<Point> p(n);
  for (auto &pi : p) cin >> pi;
  double ans = 0;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < i; j++) {
      double ma = 0, mi = 0;
      for (int k = 0; k < n; k++)
        if (k != i && k != j) {
          chmaxmin(ma, mi, cross(p[k] - p[i], p[k] - p[j]));
        }
      if (ma > 0 && mi < 0) chmax(ans, ma - mi);
    }
  cout << ans / 2 << endl;
  return 0;
}