Your security guard friend recently got a new job at a new security company. The company requires him to patrol an area of the city encompassing exactly N city blocks, but they let him choose which blocks. That is, your friend must walk the perimeter of a region whose area is exactly N blocks. Your friend is quite lazy and would like your help to find the shortest possible route that meets the requirements. The city is laid out in a square grid pattern, and is large enough that for the sake of the problem it can be considered infinite.

Input

Input will consist of a single integer N (1 ≤ N ≤ 106), the number of city blocks that must be enclosed by the route.

Output

Print the minimum perimeter that can be achieved.

Examples

Input

4


Output

8


Input

11


Output

14


Input

22


Output

20

Note

Here are some possible shapes for the examples:

<image>
import java.util.*;
public class newbar
{
    public static void main(String args[])
    {
        Scanner in=new Scanner(System.in);
        int n,size=0,i,j,lim,c=0,p=0,k=0;
        n=in.nextInt();
        for (i=1;i<=n;i++)
        {
            if (i*i>=n)
            {
                size=i;
                break;
            }
        }
        size+=2;
        int[][] a=new int[size][size];
        int m=n;
        for (i=0;i<size;i++)
        {
            for (j = 0; j < size; j++)
            {
                if (i == 0 || j == 0 || i == size - 1 || j == size - 1)
                {
                    a[i][j] = 0;
                } else
                {
                    if (m > 0)
                    {
                        a[i][j] = 1;
                        m--;
                    } else
                        a[i][j] = 0;
                }
            }
        }
        /*for (i=0;i<size;i++)
        {
            for (j=0;j<size;j++)
                System.out.print(a[i][j]+" ");
            System.out.println();
        }*/
        for (i=1;i<size-1;i++)
        {
            for (j=1;j<size-1;j++)
            {
                if (a[i][j]==1)
                {
                    k=0;
                    if (a[i][j-1]==0) k++;
                    if (a[i-1][j]==0) k++;
                    if (a[i][j+1]==0) k++;
                    if (a[i+1][j]==0) k++;
                    p=p+k;
                    //System.out.print(k+" ");
                }
            }
        }
        System.out.println(p);
    }
}
