Let's denote a function f(x) in such a way: we add 1 to x, then, while there is at least one trailing zero in the resulting number, we remove that zero. For example, 

  * f(599) = 6: 599 + 1 = 600 → 60 → 6; 
  * f(7) = 8: 7 + 1 = 8; 
  * f(9) = 1: 9 + 1 = 10 → 1; 
  * f(10099) = 101: 10099 + 1 = 10100 → 1010 → 101. 



We say that some number y is reachable from x if we can apply function f to x some (possibly zero) times so that we get y as a result. For example, 102 is reachable from 10098 because f(f(f(10098))) = f(f(10099)) = f(101) = 102; and any number is reachable from itself.

You are given a number n; your task is to count how many different numbers are reachable from n.

Input

The first line contains one integer n (1 ≤ n ≤ 10^9).

Output

Print one integer: the number of different numbers that are reachable from n.

Examples

Input


1098


Output


20


Input


10


Output


19

Note

The numbers that are reachable from 1098 are:

1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, 15, 16, 17, 18, 19, 1098, 1099.
/******************************************************************************

                            Online Java Compiler.
                Code, Compile, Run and Debug java program online.
Write your code in this editor and press "Run" button to execute it.

*******************************************************************************/
import java.io.*;
import java.util.*;
public class Codechef{
    static BufferedReader br;
    static int[] split() throws Exception
    {
        String[]cmd=br.readLine().split(" ");
        int[]res=new int[cmd.length];
        for(int i=0;i<cmd.length;i++)
        {
            res[i]=Integer.valueOf(cmd[i]);
        }
        return res;
    }
	public static void main(String[] args)throws Exception {
		br=new BufferedReader(new InputStreamReader(System.in));
		PrintWriter pr=new PrintWriter(new OutputStreamWriter(System.out));
        int t=1;
        while(t!=0)
        {
            t--;
            int n=Integer.valueOf(br.readLine());
            int ans=1;
            HashMap<Integer,Integer>map=new HashMap<>();
            map.put(n,1);
            while(true)
            {
                n+=1;
                while(n%10==0)
                {
                    n=n/10;
                }
                if(map.containsKey(n))
                break;
                ans++;
                map.put(n,1);
            }
            System.out.println(ans);
        }
	}
}
