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.
import java.util.*;
import java.io.*;
import static java.lang.System.out;


public class Main
{
    public static String remove(String s)
    {
        StringBuilder sb = new StringBuilder(s);
        while (sb.length() > 1 && sb.charAt(sb.length() - 1) == '0') 
        {
            sb.setLength(sb.length() - 1);
        }
        return sb.toString();
    }
    
    public static void main(String[] args)
    {
        Scanner sc = new Scanner(System.in);
        
        int n = sc.nextInt(), i=0, c = 0;
        Map<Integer, Integer> map = new HashMap();
        map.put(n, i); ++c;
        while(true)
        {
            n+=1;
            String s = String.valueOf(n); 
            
            if(s.charAt(s.length()-1)=='0')
            {
                s = remove(s);
                if(map.containsKey(Integer.valueOf(s))) { break; } else { map.put(Integer.valueOf(s), c); ++c; }
            }
            else
            {
                if(map.containsKey(Integer.valueOf(s))) { break; } else { map.put(Integer.valueOf(s), c); ++c; }
            }
            n = Integer.valueOf(s);
            
        }
        out.println(c);
    }
}
