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.Scanner;
//import java.lang.Math; 

public class codeforcesabc
{
	public static int fun(int a)
	{
		int c=a+1;
		while(c%10==0)
		{
			c=c/10;
		}
		return c;
	}

	public static int digits(int a)
	{
		int count=1;
		while(a/10!=0)
		{
			a=a/10;
			count++;
		}
		return count;
	}


	public static void main(String args[])
	{
		Scanner s=new Scanner(System.in);
		int n=s.nextInt();
		int d=digits(n);
		int[] arr=new int[d*10];
		//arr[0]=n;

		for(int i=0; i<d*10; i++)
		{
			arr[i]=n;
			n=fun(n);

	//		System.out.println(arr[i]);

		}
		int count=0;
		for(int i=0; i<d*10; i++)
		{
			 boolean isDistinct = false;
            for(int j=0;j<i;j++){
                if(arr[i] == arr[j]){
                    isDistinct = true;
                    break;
                }
            }
            if(!isDistinct){
            	count++;
                //System.out.print(arr[i]+" ");
            }
		}

		//int x;
		//int i=0;
	/*	int abc=n;
		//System.out.println(abc);
		while(abc%10!=0)
		{
			abc=abc+1;
			
		}
		System.out.println(abc);*/
	//	abc=d*10;
	//	System.out.println(d);
		//int x=digits(abc);

		//boolean[] visited= new boolean[d*10];
		//visited[n-1]=true;
	/*	int count=0;
		while(visited[(d*10)-1]==false)
		{
			//int x=d;
			
			visited[(d*10)-1]=true;
			n=fun(n);
			d=digits(n);
			/*if(x==d)
			{
				d=(d*10)-1;
			}
			//System.out.println(n);
			
			count=count+1;

		}*/

		System.out.println(count);


	}
}


		