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;
using ul = unsigned long long;
using pii = pair<int, int>;
using vvi = vector<vector<int>>;
using vi = vector<int>;
using vl = vector<ll>;
using pll = pair<ll, ll>;
using vsi = vector<set<int>>;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  pii pts[n];
  for (int i = 0; i < n; i++) cin >> pts[i].first >> pts[i].second;
  long double ma = 0;
  for (int i = 0; i < n; i++) {
    pii p1 = pts[i];
    for (int j = 0; j < n; j++) {
      if (j == i) continue;
      pii p2 = pts[j];
      long double a1 = 0;
      for (int k = 0; k < n; k++) {
        if (k == i || k == j) continue;
        pii p3 = pts[k];
        pii v1 = {p3.first - p1.first, p3.second - p1.second};
        pii v2 = {p2.first - p3.first, p2.second - p3.second};
        long double ta = v1.first * v2.second - v1.second * v2.first;
        if (ta > 0 && abs(ta) > a1) a1 = abs(ta);
      }
      long double a2 = 0;
      for (int k = 0; k < n; k++) {
        if (k == i || k == j) continue;
        pii p3 = pts[k];
        pii v1 = {p3.first - p1.first, p3.second - p1.second};
        pii v2 = {p2.first - p3.first, p2.second - p3.second};
        long double ta = v1.first * v2.second - v1.second * v2.first;
        if (ta < 0 && abs(ta) > a2) a2 = abs(ta);
      }
      if (min(a1, a2) > 0) ma = max(ma, (a1 + a2) / 2);
    }
  }
  cout << setprecision(10) << ma << endl;
}