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 N = 307;
const int INF = 1e9 + 7;
int n;
struct Node {
  int x, y;
  Node() {}
  Node(int _x, int _y) { x = _x, y = _y; }
} a[N];
inline int cross(const Node &p, const Node &q) { return p.x * q.y - p.y * q.x; }
int solve() {
  int ret = 0;
  for (int i = (0); i < (n); ++i) {
    for (int j = (0); j < (n); ++j) {
      if (i == j) continue;
      Node ij = Node(a[j].x - a[i].x, a[j].y - a[i].y);
      int up = -INF, dw = -INF;
      for (int k = (0); k < (n); ++k) {
        if (k == i || k == j) continue;
        Node ik = Node(a[k].x - a[i].x, a[k].y - a[i].y);
        int cr = cross(ik, ij);
        if (cr < 0) {
          up = max(up, abs(cr));
        } else {
          dw = max(dw, abs(cr));
        }
      }
      ret = max(ret, up + dw);
    }
  }
  return ret;
}
int main() {
  scanf("%d", &n);
  for (int i = (0); i < (n); ++i) scanf("%d%d", &a[i].x, &a[i].y);
  printf("%.15f", solve() / 2.);
  return 0;
}