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;
pair<long long, long long> operator-(pair<long long, long long> a,
                                     pair<long long, long long> b) {
  return {a.first - b.first, a.second - b.second};
}
long long operator%(pair<long long, long long> a,
                    pair<long long, long long> b) {
  return a.first * b.second - a.second * b.first;
}
long long operator*(pair<long long, long long> a,
                    pair<long long, long long> b) {
  return a.first * b.first + a.second * b.second;
}
pair<long long, long long> operator+(pair<long long, long long> a,
                                     pair<long long, long long> b) {
  return {a.first + b.first, a.second + b.second};
}
double operator~(pair<long long, long long> a) {
  return hypot(a.first, a.second);
}
bool angle_cmp(pair<long long, long long> a, pair<long long, long long> b) {
  if (a % b == 0) return a * a < b * b;
  return a % b > 0;
}
vector<pair<long long, long long> > convexHull(
    vector<pair<long long, long long> > a) {
  pair<long long, long long> base = *min_element(a.begin(), a.end());
  for (pair<long long, long long>& p : a) {
    p = p - base;
  }
  sort(a.begin(), a.end(), angle_cmp);
  vector<pair<long long, long long> > hull;
  for (pair<long long, long long> p : a) {
    while (hull.size() >= 2 &&
           (p - hull.end()[-2]) % (hull.end()[-1] - hull.end()[-2]) >= 0) {
      hull.pop_back();
    }
    hull.push_back(p);
  }
  return hull;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  vector<pair<long long, long long> > p(n);
  for (auto& first : p) cin >> first.first >> first.second;
  long long best = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (i == j) continue;
      long long mi = -1, ma = -1;
      for (int k = 0; k < n; k++) {
        if (k == i || k == j) continue;
        pair<long long, long long> a = p[i], b = p[j], c = p[k];
        if ((a - b) % (a - c) > 0) {
          mi = max(mi, abs(abs(a % b + b % c + c % a)));
        } else {
          ma = max(ma, abs(abs(a % b + b % c + c % a)));
        }
      }
      if (mi > -1 && ma > -1) best = max(best, mi + ma);
    }
  }
  cout << setprecision(20) << best / 2. << endl;
}