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;
int CCW(pair<int, int> A, pair<int, int> B, pair<int, int> C) {
  int u1 = B.first - A.first;
  int v1 = B.second - A.second;
  int u2 = C.first - A.first;
  int v2 = C.second - A.second;
  if (u1 * v2 - u2 * v1 > 0) return 1;
  return -1;
}
int abs(int n) {
  if (n >= 0) return n;
  return -n;
}
int area(pair<int, int> A, pair<int, int> B, pair<int, int> C) {
  int res = (B.first - A.first) * (B.second + A.second) +
            (C.first - B.first) * (C.second + B.second) +
            (A.first - C.first) * (A.second + C.second);
  return abs(res);
}
void maximize(int &a, int b) {
  if (a < b) a = b;
}
pair<int, int> p[320];
int n;
int main() {
  scanf("%d", &n);
  int ans = 0;
  for (int i = 1; i <= n; i++) scanf("%d %d", &p[i].first, &p[i].second);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) {
      int left = 0;
      int right = 0;
      for (int k = 1; k <= n; k++) {
        if (k == i || k == j) continue;
        if (CCW(p[i], p[j], p[k]) < 0)
          maximize(left, area(p[i], p[j], p[k]));
        else
          maximize(right, area(p[i], p[j], p[k]));
      }
      if (left > 0 && right > 0) maximize(ans, left + right);
    }
  printf("%lf", (double)ans / 2);
}