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;
double EPS = 1e-9;
int cmp(double a, double b = 0) {
  return a + EPS < b ? -1 : a - EPS > b ? 1 : 0;
}
struct Point {
  int x, y;
  int id;
  Point() {}
  Point(int x, int y) : x(x), y(y) {}
  bool operator<(const Point &p) const {
    if (y == p.y) return x < p.x;
    return y < p.y;
  }
  int operator%(Point p) { return x * p.y - y * p.x; }
  Point operator-(Point p) { return Point(x - p.x, y - p.y); }
  int operator*(Point p) { return x * p.x + y * p.y; }
};
int ccw(Point a, Point b, Point c) { return (c - a) % (b - a); }
Point p[1005];
Point h[2005];
void hull(int &n) {
  int k = 0;
  for (int i = 0; i < n; i++) {
    while (k >= 2 && ccw(h[k - 2], h[k - 1], p[i]) <= 0) k--;
    h[k++] = p[i];
  }
  int t = k + 1;
  for (int i = n - 2; i >= 0; i--) {
    while (k >= t && ccw(h[k - 2], h[k - 1], p[i]) <= 0) k--;
    h[k++] = p[i];
  }
  n = k - 1;
}
int area(int i, int j, int k) {
  return fabs(h[i] % h[j] + h[j] % h[k] + h[k] % h[i]);
}
int solve(int n) {
  int no = n;
  sort(p, p + n);
  int res = 0.0;
  hull(n);
  if (n >= 4) {
    for (int i = 0; i < n; i++) {
      int q = i, r = (i + 1) % n, j = (i + 1) % n;
      for (int k = 1; k < n; k++, j = (j + 1) % n) {
        while (area(i, j, (q + 1) % n) > area(i, j, q)) q = (q + 1) % n;
        while (area(i, j, (r + 1) % n) > area(i, j, r)) r = (r + 1) % n;
        res = max(res, area(i, j, q) + area(i, r, j));
      }
    }
  } else {
    for (int i = 0; i < n; i++) {
      int a = 0, b = 1;
      if (i == 0)
        a = 1, b = 2;
      else if (i == 1)
        a = 0, b = 2;
      Point pt = (h[a] - h[b]);
      for (int j = 0; j < no; j++)
        if (p[j].id != h[a].id && p[j].id != h[b].id && p[j].id != h[i].id)
          res = max(res, abs(pt % (h[i] - p[j])));
    }
  }
  return res;
}
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> p[i].x >> p[i].y;
    p[i].id = i;
  }
  double res = solve(n) / 2.;
  printf("%.9lf\n", res);
  return 0;
}