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 dt {
  int a, b, c;
  dt(){};
  dt(int A, int B, int C) {
    a = A;
    b = B;
    c = C;
  }
  dt(pair<int, int> p1, pair<int, int> p2) {
    a = p1.second - p2.second;
    b = p2.first - p1.first;
    c = p1.first * p2.second - p2.first * p1.second;
  }
  int vt(pair<int, int> p) {
    if ((a * p.first + b * p.second + c) > 0) return 1;
    if ((a * p.first + b * p.second + c) < 0) return -1;
    return 0;
  }
};
const int maxn = 301;
int n;
pair<int, int> a[maxn];
double res = -1e9;
double dis(pair<int, int> p1, pair<int, int> p2) {
  return sqrt((p1.first - p2.first) * (p1.first - p2.first) +
              (p1.second - p2.second) * (p1.second - p2.second));
}
double area(pair<int, int> p1, pair<int, int> p2, pair<int, int> p3) {
  double p12, p23, p13;
  p12 = dis(p1, p2);
  p23 = dis(p2, p3);
  p13 = dis(p1, p3);
  double p = (p12 + p23 + p13) / 2.0;
  return sqrt(p * (p - p12) * (p - p23) * (p - p13));
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i].first >> a[i].second;
  int ck;
  double rescp, resnp;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++)
      if (i != j) {
        dt ij(a[i], a[j]);
        rescp = -1;
        resnp = -1;
        for (int k = 1; k <= n; k++) {
          ck = ij.vt(a[k]);
          if (ck == 1)
            rescp = max(rescp, area(a[i], a[j], a[k]));
          else if (ck == -1)
            resnp = max(resnp, area(a[i], a[j], a[k]));
        }
        if (rescp != -1 && resnp != -1) res = max(res, rescp + resnp);
      }
  cout << setprecision(8) << res;
}