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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.HashSet;
import java.util.Set;

public class Div2553 {

    public static void main(String[] args) throws IOException {

        BufferedReader gi = new BufferedReader(new InputStreamReader(System.in));

        String[] s1 = gi.readLine().split(" ");
        int n = Integer.parseInt(s1[0]);
        
         int i =1;
         String  numZeroes ;
         Set<Integer> s = new HashSet<Integer>();
         n++;
          numZeroes=Integer.toString(n).replaceAll("0+$", "");
          
        while(!s.contains(n)){
              s.add(n);
            
             n=Integer.parseInt(numZeroes)+1;
           numZeroes=Integer.toString(n).replaceAll("0+$", "");
           
            
           
            i++;
        }
        System.out.println(i-1);
           
    }

    static int[] parseArray(BufferedReader gi) throws IOException {
        String[] line = gi.readLine().trim().split(" ");
        int[] rez = new int[line.length];
        for (int k = 0; k < line.length; k++) {
            rez[k] = Integer.parseInt(line[k]);
        }
        return rez;
    }

     static int findTrailingZeros(int n) 
    { 
        // Initialize result 
        int count = 0; 
  
        // Keep dividing n by powers  
        // of 5 and update count 
        for (int i = 5; n / i >= 1; i *= 5) 
            count += n / i; 
  
        return count; 
    } 
  
}