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 datos {
  double x, y;
};
int n;
datos puntos[302];
double sol;
double cruz(int o, int a, int b) {
  datos uno, dos;
  uno.x = puntos[a].x - puntos[o].x;
  uno.y = puntos[a].y - puntos[o].y;
  dos.x = puntos[b].x - puntos[o].x;
  dos.y = puntos[b].y - puntos[o].y;
  double res = (uno.x * dos.y) - (dos.x * uno.y);
  return res;
}
void lineas() {
  for (int i = 1; i <= n; i++) {
    for (int j = i + 1; j <= n; j++) {
      double area_izq = 0.0;
      double area_der = 0.0;
      int ni = 0, nd = 0;
      for (int k = 1; k <= n; k++) {
        if (k != i && k != j) {
          double lado = cruz(i, j, k);
          if (lado > 0) {
            area_izq = max(area_izq, abs(cruz(k, i, j)) / 2.0);
            ni++;
          } else {
            area_der = max(area_der, abs(cruz(k, i, j)) / 2.0);
            nd++;
          }
        }
      }
      if (ni && nd) sol = max(sol, area_izq + area_der);
    }
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) cin >> puntos[i].x >> puntos[i].y;
  lineas();
  printf("%.9lf\n", sol);
  return 0;
}