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;
struct point {
  int x, y;
};
point operator-(point a, point b) {
  point res = {a.x - b.x, a.y - b.y};
  return res;
}
int operator*(point a, point b) { return a.x * b.y - a.y * b.x; }
vector<point> arr;
int s(int i, int j, int q) {
  point a = arr[i] - arr[q], b = arr[j] - arr[q];
  return a * b;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  int n;
  cin >> n;
  arr.resize(n);
  for (auto& i : arr) cin >> i.x >> i.y;
  int ans = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (i == j) continue;
      bool f1 = 0, f2 = 0;
      int temp = 0, sum = 0;
      for (int q = 0; q < n; q++) {
        if (i == q || j == q) continue;
        point a = arr[i] - arr[q], b = arr[j] - arr[q];
        int s = a * b;
        if (s < 0) continue;
        f1 = 1;
        s = abs(s);
        if (s > temp) temp = s;
      }
      sum += temp;
      temp = 0;
      for (int q = 0; q < n; q++) {
        if (i == q || j == q) continue;
        point a = arr[i] - arr[q], b = arr[j] - arr[q];
        int s = a * b;
        if (s > 0) continue;
        f2 = 1;
        s = abs(s);
        if (s > temp) temp = s;
      }
      sum += temp;
      if (f1 && f2) ans = max(ans, sum);
    }
  }
  cout << fixed << setprecision(10) << (double)ans * 0.5;
  return 0;
}