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 timer {
  time_t start;
  timer() { start = clock(); }
  ~timer() {
    cerr << 1. * (clock() - start) / CLOCKS_PER_SEC << " secs" << endl;
  }
};
int n, x[310], y[310], dp[5][310], start;
int f(int num, int cur) {
  int& res = dp[num][cur];
  if (res != -1070000000LL * 2) return res;
  if (num == 4) return res = cur == start ? 0 : -1070000000LL;
  res = -1070000000LL;
  for (int i = 0; i < (n); i++)
    if (i != cur) {
      res = max(res, x[cur] * y[i] - x[i] * y[cur] + f(num + 1, i));
    }
  return res;
}
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  cin >> n;
  for (int i = 0; i < (n); i++) cin >> x[i] >> y[i];
  int ans = 0;
  for (int i = 0; i < (n); i++) {
    for (int j = 0; j < (5); j++)
      for (int k = 0; k < (n); k++) dp[j][k] = -1070000000LL * 2;
    start = i;
    ans = max(ans, f(0, i));
  }
  cout << setprecision(16) << ans / 2.;
}