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.util.*;
public class ReachableNumbers
{
	static long cnt;
	public static void main(String args[])throws IOException
	{
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter pw = new PrintWriter(System.out);
	/*	long ar[] = new long[1000001];
		long mod = (long)(Math.pow(10 , 9) + 7);
		for(int i = 1 ; i <= 1000000 ; i++)
		{
			ar[i] = (((ar[i - 1] + i) % mod) + ((ar[i - 1] * i)) % mod) % mod;
		}*/
		long n = Long.parseLong(br.readLine().trim());
		LinkedHashSet<Long> set = new LinkedHashSet<Long>();
		set.add(n);
		while(true)
		{
			n = getf(n);
			if(!set.add(n))
			{
				break;
			}
		}
		pw.println(set.size());
		pw.flush();
		
	}
	public static long getf(long n)
	{
		n += 1;
		if(n % 10 != 0)
		{
			return n;
		}
		while(true)
		{
			n /= 10;
			if(n % 10 != 0)
			{
				break;
			}
		}
		return n;
	}
}

