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 Point {
  int x, y;
  Point() {}
  Point(int xx, int yy) : x(xx), y(yy) {}
  friend bool operator<(const Point &a, const Point &b) {
    return a.x < b.x || a.x == b.x && a.y == b.y;
  }
  friend bool operator==(const Point &a, const Point &b) {
    return a.x == b.x && a.y == b.y;
  }
};
Point operator-(Point a, Point b) { return Point(a.x - b.x, a.y - b.y); }
struct polygon_convex {
  vector<Point> P;
  polygon_convex(int Size = 0) { P.resize(Size); }
};
int Cross(Point a, Point b) { return a.x * b.y - a.y * b.x; }
polygon_convex convex_hull(vector<Point> a) {
  polygon_convex res(2 * a.size() + 5);
  sort(a.begin(), a.end());
  a.erase(unique(a.begin(), a.end()), a.end());
  int m = 0;
  for (int i = 0; i < a.size(); ++i) {
    while (m > 1 &&
           Cross(res.P[m - 1] - res.P[m - 2], a[i] - res.P[m - 2]) <= 0)
      --m;
    res.P[m++] = a[i];
  }
  int k = m;
  for (int i = int(a.size()) - 2; i >= 0; --i) {
    while (m > k &&
           Cross(res.P[m - 1] - res.P[m - 2], a[i] - res.P[m - 2]) <= 0)
      --m;
    res.P[m++] = a[i];
  }
  res.P.resize(m);
  if (a.size() > 1) res.P.resize(m - 1);
  return res;
}
int main() {
  vector<Point> a;
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    Point p;
    scanf("%d%d", &p.x, &p.y);
    a.push_back(p);
  }
  vector<Point> p = convex_hull(a).P;
  int m = p.size();
  int max = 0;
  if (m == 3) {
    int ans = Cross(p[1] - p[0], p[2] - p[0]);
    int min = 0x7fffffff;
    for (int i = 0; i < m; i++) {
      for (int j = 0; j < n; j++) {
        int area = Cross(p[(i + 1) % m] - p[i], a[j] - p[i]);
        if (area > 0 && area < min) {
          min = area;
        }
      }
    }
    printf("%.6lf\n", fabs(ans - min) / 2.0);
  } else {
    for (int i = 0; i < m; i++) {
      for (int j = i + 1; j < m; j++) {
        for (int k = j + 1; k < m; k++) {
          for (int l = k + 1; l < m; l++) {
            int area = Cross(p[j] - p[i], p[k] - p[i]) +
                       Cross(p[k] - p[i], p[l] - p[i]);
            if (max < area) {
              max = area;
            }
          }
        }
      }
    }
    printf("%.6lf\n", fabs(max / 2.0));
  }
  return 0;
}