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>
const int dx[4] = {0, 0, 0, 0};
using namespace std;
struct Point {
  double x, y;
  Point(double x = 0, double y = 0) : x(x), y(y) {}
  bool operator<(const Point b) const {
    if (b.x == x) return y < b.y;
    return x < b.x;
  }
} p[310], ch[310];
Point operator-(Point A, Point B) { return Point(A.x - B.x, A.y - B.y); }
bool operator!=(Point A, Point B) {
  if (A.x != B.x || A.y != B.y)
    return true;
  else
    return false;
}
double Cross(Point A, Point B) { return A.x * B.y - A.y * B.x; }
int ConvexHull(int n) {
  int i, top, t;
  sort(p, p + n);
  top = 0;
  for (i = 0; i < n; i++) {
    while (top > 1 && Cross(ch[top - 1] - ch[top - 2], p[i] - ch[top - 2]) <= 0)
      top--;
    ch[top++] = p[i];
  }
  t = top;
  for (i = n - 2; i >= 0; i--) {
    while (top > t && Cross(ch[top - 1] - ch[top - 2], p[i] - ch[top - 2]) <= 0)
      top--;
    ch[top++] = p[i];
  }
  if (n > 1) top--;
  return top;
}
double quadarea(int n) {
  int i, j, k, l;
  double ans = 0, area;
  for (i = 0; i <= n - 4; i++) {
    for (j = i + 1; j <= n - 3; j++) {
      for (k = j + 1; k <= n - 2; k++) {
        for (l = k + 1; l < n; l++) {
          area = Cross(ch[j] - ch[i], ch[k] - ch[i]) +
                 Cross(ch[k] - ch[i], ch[l] - ch[i]);
          if (ans < area) ans = area;
        }
      }
    }
  }
  return ans / 2.0;
}
double max(double a, double b, double c) {
  a = (a > b ? a : b);
  a = (a > c ? a : c);
  return a;
}
double qquadarea(int n) {
  int i;
  double ans = 0, temp1, temp2, temp3;
  for (i = 0; i < n; i++) {
    if (p[i] != ch[0] && p[i] != ch[1] && p[i] != ch[2]) {
      temp1 = Cross(ch[1] - ch[0], p[i] - ch[0]) +
              Cross(p[i] - ch[0], ch[2] - ch[0]);
      temp2 = Cross(ch[0] - ch[1], p[i] - ch[1]) +
              Cross(p[i] - ch[1], ch[2] - ch[1]);
      temp3 = Cross(ch[0] - ch[2], p[i] - ch[2]) +
              Cross(p[i] - ch[2], ch[1] - ch[2]);
      temp1 = max(temp1, temp2, temp3);
    }
    if (ans < temp1) ans = temp1;
  }
  return ans / 2.0;
}
int main(int argc, char** argv) {
  int n, dd, i;
  double area;
  while (scanf("%d", &n) != EOF) {
    for (i = 0; i < n; i++) scanf("%lf%lf", &p[i].x, &p[i].y);
    dd = ConvexHull(n);
    if (dd >= 4)
      area = quadarea(dd);
    else if (dd >= 1 && dd < 3)
      area = 0;
    else if (dd == 3)
      area = qquadarea(n);
    printf("%lf\n", area);
  }
  return 0;
}