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 <typename T>
inline void _abs(T &first) {
  if (first < 0) first = -first;
};
const int N = 1606;
int n;
pair<long long, long long> a[N];
void inp() {
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i].first >> a[i].second;
}
long long dientich(pair<long long, long long> a, pair<long long, long long> b,
                   pair<long long, long long> c) {
  long long t1 = 1ll * (b.first - a.first) * (b.second + a.second);
  long long t2 = 1ll * (c.first - b.first) * (c.second + b.second);
  long long t3 = 1ll * (a.first - c.first) * (a.second + c.second);
  return t1 + t2 + t3;
}
void solve() {
  double res = 0;
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= n; ++j) {
      double max1 = 0, max2 = 0;
      for (int k = 1; k <= n; ++k) {
        long long S = dientich(a[i], a[j], a[k]);
        if (S < 0)
          max1 = max(max1, (double)-S / 2);
        else
          max2 = max(max2, (double)S / 2);
      }
      if (max1 && max2) res = max(res, max1 + max2);
    }
  }
  cout << (int)res << (res - (int)res ? ".5" : ".0");
}
int main() {
  ios ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  inp();
  solve();
  return 0;
}