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.io.*;
import java.util.*;
public class div429A
{
	BufferedReader in;
	PrintWriter ob;
	StringTokenizer st;
	public static void main(String[] args) throws IOException {
		new div429A().run();
	}
	void run() throws IOException {
		//in=new BufferedReader(new FileReader("input.txt"));
		in=new BufferedReader(new InputStreamReader(System.in));
		ob=new PrintWriter(System.out);
		solve();
		ob.flush();
	}
	void solve() throws IOException {
		int n = ni();	
		long ans = Long.MAX_VALUE;
		for(int i=1 ; (i*i)<=5*1L*n ; i++) {
			for(int j=1 ; (j*j)<=5*1L*n ; j++) {
				if(i*1L*j >= n ) {
					ans = Math.min(ans , 2*1L*(i+j));
				}
			}
		}
		ob.println(ans);
	}
	String ns() throws IOException {
		return nextToken();
	}
	long nl() throws IOException {
		return Long.parseLong(nextToken());
	}
	int ni() throws IOException {
		return Integer.parseInt(nextToken());
	}
	double nd() throws IOException {
		return Double.parseDouble(nextToken());
	} 
	String nextToken() throws IOException {
		if(st==null || !st.hasMoreTokens())
			st=new StringTokenizer(in.readLine());
		return st.nextToken();
	}
	int[] nia(int start,int b) throws IOException {
		int a[]=new int[b];
		for(int i=start;i<b;i++)
			a[i]=ni();
		return a;
	}
	long[] nla(int start,int n) throws IOException {
		long a[]=new long[n];
		for (int i=start; i<n ;i++ ) {
			a[i]=nl();
		}
		return a;
	}
}