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 double eps = 1e-9;
struct point {
  double x, y;
  point() { x = y = 0.0; }
  point(double _x, double _y) { x = _x, y = _y; }
};
struct vec {
  double x, y;
  vec() { x = y = 0.0; }
  vec(point a, point b) { x = b.x - a.x, y = b.y - a.y; }
  double operator^(const vec& rhs) { return x * rhs.y - y * rhs.x; }
};
double cross(point p, point q, point r) {
  vec pq(p, q), pr(p, r);
  return (pq ^ pr);
}
vector<point> v;
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    v.emplace_back(x, y);
  }
  double ans = -100;
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      point p = v[i], q = v[j];
      double area1 = -100, area2 = -100;
      for (int k = 0; k < n; k++) {
        if (k == i or k == j) continue;
        point r = v[k];
        double crs = cross(p, q, r) / 2.0;
        if (crs > 0) area1 = max(abs(crs), area1);
        if (crs < 0) area2 = max(abs(crs), area2);
      }
      if (area1 < 0) continue;
      if (area2 < 0) continue;
      ans = max(ans, area1 + area2);
    }
  }
  printf("%0.6lf\n", ans);
}