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;
using ll = long long;
struct point {
  ll x, y;
};
struct line {
  ll a, b, c;
  line(point A, point B) {
    a = B.y - A.y;
    b = A.x - B.x;
    c = -(a * A.x + b * A.y);
  }
  bool up(point P) { return a * P.x + b * P.y + c > 0; }
};
int n;
vector<point> p;
vector<line> l;
ll S(point a, point b, point c) {
  return abs((b.x - a.x) * (c.y - a.y) - (c.x - a.x) * (b.y - a.y));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    ll x, y;
    cin >> x >> y;
    p.push_back({x, y});
  }
  ll ans = 0;
  for (int i = 0; i < n; ++i) {
    for (int j = i + 1; j < n; ++j) {
      ll up = -1e9 - 7, down = -1e9 - 7;
      line li = line(p[i], p[j]);
      for (int k = 0; k < i; ++k) {
        if (li.up(p[k]))
          up = max(up, S(p[i], p[j], p[k]));
        else
          down = max(down, S(p[i], p[j], p[k]));
      }
      for (int k = i + 1; k < j; ++k) {
        if (li.up(p[k]))
          up = max(up, S(p[i], p[j], p[k]));
        else
          down = max(down, S(p[i], p[j], p[k]));
      }
      for (int k = j + 1; k < n; ++k) {
        if (li.up(p[k]))
          up = max(up, S(p[i], p[j], p[k]));
        else
          down = max(down, S(p[i], p[j], p[k]));
      }
      ans = max(ans, up + down);
    }
  }
  cout << ans / 2;
  if (ans % 2) cout << ".5";
}