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;
  bool operator<(const point& a) const {
    if (a.x == x) return a.y < y;
    return a.x < x;
  }
};
point operator-(point a, point b) {
  point temp;
  temp.x = a.x - b.x;
  temp.y = a.y - b.y;
  return temp;
}
bool operator==(point a, point b) { return a.x == b.x && a.y == b.y; }
double cross(point a, point b) { return a.x * b.y - a.y * b.x; }
int convexhull(point* p, int n, point* ch) {
  int m = 0;
  for (int i = 0; i < n; i++) {
    while (m > 1 && cross(ch[m - 1] - ch[m - 2], p[i] - ch[m - 2]) <= 0) m--;
    ch[m++] = p[i];
  }
  int k = m;
  for (int i = n - 2; i >= 0; i--) {
    while (m > k && cross(ch[m - 1] - ch[m - 2], p[i] - ch[m - 2]) <= 0) m--;
    ch[m++] = p[i];
  }
  if (n > 1) m--;
  return m;
}
int main() {
  int n;
  point num[3050];
  point ch[3050];
  double ans[3050];
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> num[i].x >> num[i].y;
  }
  sort(num, num + n);
  int m;
  m = convexhull(num, n, ch);
  double area = 0;
  double maxarea = 0;
  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 d = k + 1; d < m; d++) {
          area = cross(ch[i], ch[j]);
          area += cross(ch[j], ch[k]);
          area += cross(ch[k], ch[d]);
          area += cross(ch[d], ch[i]);
          if (area > maxarea) maxarea = area;
        }
      }
    }
  }
  if (m == 3) {
    maxarea = 0;
    area = cross(ch[0], ch[1]) + cross(ch[1], ch[2]) + cross(ch[2], ch[0]);
    for (int i = 0; i < n; i++) {
      if (num[i] == ch[0] || num[i] == ch[1] || num[i] == ch[2]) continue;
      double temp =
          cross(num[i], ch[1]) + cross(ch[1], ch[2]) + cross(ch[2], num[i]);
      temp = area - temp;
      if (temp > maxarea) maxarea = temp;
    }
  }
  printf("%.6lf\n", maxarea / 2);
  return 0;
}