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;
const int maxn = 400;
long long n;
long long dp[maxn][maxn][2];
pair<long long, long long> p[maxn];
inline void init() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> p[i].first >> p[i].second;
}
inline int f(long long x) { return x < 0; }
inline long long cross(pair<long long, long long> p1,
                       pair<long long, long long> p2,
                       pair<long long, long long> p3) {
  long long x22 = p3.first - p1.first, x11 = p2.first - p1.first,
            y22 = p3.second - p1.second, y11 = p2.second - p1.second;
  return x11 * y22 - x22 * y11;
}
inline void solve() {
  for (int i = 0; i < n; i++)
    for (int j = i + 1; j < n; j++)
      for (int k = 0; k < n; k++)
        if (i != k && j != k) {
          long long hlp = cross(p[i], p[j], p[k]);
          dp[i][j][f(hlp)] = max(dp[i][j][f(hlp)], abs(hlp));
        }
  long long ans = 0;
  for (int i = 0; i < n; i++)
    for (int j = i + 1; j < n; j++)
      if (dp[i][j][0] != 0 && dp[i][j][1] != 0)
        ans = max(ans, dp[i][j][0] + dp[i][j][1]);
  cout << fixed << setprecision(10) << ans / (double)2 << endl;
}
int main() {
  init();
  solve();
  return 0;
}