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;
template <typename A, typename B>
ostream &operator<<(ostream &os, const pair<A, B> &p) {
  return os << '(' << p.first << ", " << p.second << ')';
}
template <typename T_container, typename T = typename enable_if<
                                    !is_same<T_container, string>::value,
                                    typename T_container::value_type>::type>
ostream &operator<<(ostream &os, const T_container &v) {
  os << '{';
  string sep;
  for (const T &x : v) os << sep << x, sep = ", ";
  return os << '}';
}
void dbg_out() { cerr << endl; }
template <typename Head, typename... Tail>
void dbg_out(Head H, Tail... T) {
  cerr << ' ' << H;
  dbg_out(T...);
}
using vi = vector<long long>;
using pii = pair<long long, long long>;
using vp = vector<pii>;
using vv = vector<vector<long long>>;
const long long MOD = (long long)1e9 + 7;
const long long MOD2 = (long long)998244353;
const long long inf = (long long)1e18 + 1;
long long powb(long long a, long long n) {
  long long res = 1;
  while (n > 0) {
    if (n & 1) res = res * a;
    a = a * a;
    n >>= 1;
  }
  return res;
}
long long powb(long long a, long long n, long long m) {
  a %= m;
  long long res = 1;
  while (n > 0) {
    if (n & 1) res = (res * a) % m;
    a = (a * a) % m;
    n >>= 1;
  }
  return res;
}
long double area(long long i, long long A, long long B, vi &x, vi &y) {
  long double ans = 0.5;
  ans *= (x[i] - x[A]) * (y[B] - y[A]) - (y[i] - y[A]) * (x[B] - x[A]);
  return ans;
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  vi x(n);
  vi y(n);
  for (long long i = 0; i < n; i++) {
    cin >> x[i] >> y[i];
  }
  long double mx = -inf;
  for (long long A = 0; A < n; A++) {
    for (long long B = A + 1; B < n; B++) {
      long double mx_right = 0;
      long double mx_left = 0;
      for (long long i = 0; i < n; i++) {
        long double curr = area(i, A, B, x, y);
        if (curr > 0) mx_right = max(mx_right, curr);
        if (curr < 0) mx_left = max(mx_left, abs(curr));
      }
      if (mx_left == 0 || mx_right == 0) continue;
      mx = max(mx_left + mx_right, mx);
    }
  }
  cout << fixed << showpoint;
  cout << setprecision(10);
  cout << mx << "\n";
  return 0;
}