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.
/*
 * To change this license header, choose License Headers in Project Properties.
 * To change this template file, choose Tools | Templates
 * and open the template in the editor.
 */
//package Round555;

import java.util.HashSet;
import java.util.Scanner;
import java.util.Set;

/**
 *
 * @author Handsome
 */
public class ReachableNumbers {
    public static void main(String[] args) {
        long n;
        Scanner inp = new Scanner(System.in);
        
        n = inp.nextLong();
        
        System.out.println(countReachableNumbers(n));
    }

    private static int countReachableNumbers(long n) {
        int res = 0;
        Set<Long> set = new HashSet<>();
        
        while( !set.contains(n) ){
            set.add(n);
            n = f(n);
            res++;
        }
        
        return res;
    }

    private static long f(long n) {
        n++;
        while(n%10 == 0) n/=10;
        return n;
    }
    
    
}
