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 k, m;
pair<double, double> a[305];
inline double get_dis(pair<double, double> a, pair<double, double> b) {
  double dis;
  dis = sqrt((a.first - b.first) * (a.first - b.first) +
             (a.second - b.second) * (a.second - b.second));
  return dis;
}
inline bool judge(pair<double, double> x, pair<double, double> y,
                  pair<double, double> z) {
  return (z.first - x.first) * (y.second - x.second) / (y.first - x.first) +
             x.second <
         z.second;
}
inline double hailun(pair<double, double> x, pair<double, double> y,
                     pair<double, double> z) {
  double a = get_dis(x, y), b = get_dis(y, z), c = get_dis(x, z),
         pi = (a + b + c) / 2;
  return sqrt(pi * (pi - a) * (pi - b) * (pi - c));
}
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i].first >> a[i].second;
  }
  sort(a + 1, a + 1 + n);
  double ans = 0;
  for (int i = 1; i < n; i++) {
    for (int j = i + 1; j <= n; j++) {
      double ans1 = 0, ans2 = 0;
      for (int m = 1; m <= n; m++) {
        if (m != i && m != j) {
          if (judge(a[i], a[j], a[m]))
            ans1 = max(ans1, hailun(a[i], a[j], a[m]));
          else {
            ans2 = max(ans2, hailun(a[i], a[j], a[m]));
          }
        }
      }
      if (ans1 != 0 && ans2 != 0) ans = max(ans, ans1 + ans2);
    }
  }
  printf("%llf", ans);
  return 0;
}