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 JAVA):

Solution:
import java.util.*;

public class MaximalAreaQuadrilateral{
   public static void main(String[] args){
      Scanner sc = new Scanner(System.in);
      int n = sc.nextInt();
      int[] xs = new int[n];
      int[] ys = new int[n];
      for(int k = 0; k < n; k++){
         xs[k] = sc.nextInt();
         ys[k] = sc.nextInt();
      }
      long ans = 0;
      for(int k = 0; k < n; k++){
         for(int j = k+1; j < n; j++){
            long pos = -1000000;
            long neg = 1000000;
            int x1, x2, y1, y2;
            for(int i = 0; i <n;i++){
               x1 = xs[k]-xs[i];
               x2 = xs[j]-xs[i];
               y1 = ys[k]-ys[i];
               y2 = ys[j]-ys[i];
               if(i!=j && i !=k){
                  if(x1*y2-x2*y1 > 0)
                     pos = Math.max(pos, x1*y2-x2*y1);
                  if(x1*y2-x2*y1 < 0)
                     neg = Math.min(neg, x1*y2-x2*y1);
               }
            }
            ans = Math.max(ans, pos-neg);
         }
      }
      System.out.println(ans*.5);
   }
}