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.*;
import java.lang.*;
import java.util.*;

public class A1157 {
    public static void main(String[] args) throws IOException{
        StringBuffer ans = new StringBuffer();
        StringTokenizer st;
        BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
        st = new StringTokenizer(f.readLine());
        int n = Integer.parseInt(st.nextToken());
        HashSet<Integer> hash = new HashSet<>();
        int op = 1;
        hash.add(n);
        while (true){
            int r = f(n);
            //System.out.println(n + " " + r);
            if(hash.contains(r)){
                break;
            }
            hash.add(r);
            n = r;
            op++;
        }
        f.close();



        System.out.println(op);
    }
    public static int f(int n){
        n++;
        String r = Integer.toString(n);
        while (r.charAt(r.length()-1) == '0'){
            n/=10;
            r = Integer.toString(n);
        }
        return n;
    }

}