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;
double max(double a, double b) { return (a > b) ? a : b; }
double cross(pair<int, int> a, pair<int, int> b, pair<int, int> c) {
  return ((double)(b.first - a.first) * (c.second - b.second) -
          (b.second - a.second) * (c.first - b.first)) *
         0.5;
}
int main() {
  long long int n;
  cin >> n;
  pair<int, int> p[n];
  for (int i = 0; i < n; i++) {
    cin >> p[i].first >> p[i].second;
  }
  double ans = 0;
  double cminus = 0, cplus = 0;
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      cminus = -1;
      cplus = -1;
      for (int k = 0; k < n; k++) {
        if (i != k && j != k) {
          if (cross(p[i], p[j], p[k]) < 0) {
            if (cminus < -cross(p[i], p[j], p[k])) {
              cminus = -cross(p[i], p[j], p[k]);
            }
          } else {
            if (cplus < cross(p[i], p[j], p[k])) {
              cplus = cross(p[i], p[j], p[k]);
            }
          }
        }
      }
      if (cminus >= 0 && cplus >= 0) ans = max(ans, cminus + cplus);
    }
  }
  printf("%lf", ans);
}