Problem: You are given a positive integer n.

Let S(x) be sum of digits in base 10 representation of x, for example, S(123) = 1 + 2 + 3 = 6, S(0) = 0.

Your task is to find two integers a, b, such that 0 ≤ a, b ≤ n, a + b = n and S(a) + S(b) is the largest possible among all such pairs.

Input

The only line of input contains an integer n (1 ≤ n ≤ 10^{12}).

Output

Print largest S(a) + S(b) among all pairs of integers a, b, such that 0 ≤ a, b ≤ n and a + b = n.

Examples

Input

35


Output

17


Input

10000000000


Output

91

Note

In the first example, you can choose, for example, a = 17 and b = 18, so that S(17) + S(18) = 1 + 7 + 1 + 8 = 17. It can be shown that it is impossible to get a larger answer.

In the second test example, you can choose, for example, a = 5000000001 and b = 4999999999, with S(5000000001) + S(4999999999) = 91. It can be shown that it is impossible to get a larger answer.

Solution (in JAVA):

Solution:
import java.util.*;
public class ph{
	//hi
	static long sum(long a){
		long s=0;
		while(a>0){
			long b=a%10;
			 s=s+b;
			a=a/10;
		}
		return s;
	}
	public static void main(String[] args) {
		Scanner sc=new Scanner(System.in);
		long n=sc.nextLong();
		long b=n,c=0,u=0,s=0,i;
		while(b>0){
			u=b%10;
			b=b/10;
			c++;
		}
		//System.out.println(c+" "+u);
		long g[]=new long[(int)c];
		for(i=c-1;i>=1;i--){
			g[(int)i]=9;
		}
		g[0]=u-1;
		/*for(i=0;i<c;i++){
			System.out.print(g[(int)i]);
		}*/
		//System.out.println();
		s=0;
		for(i=0;i<c;i++){
			s=s+g[(int)i]*(long)Math.pow(10,c-i-1);
			//System.out.println(s);
		}
		System.out.println(sum(s)+sum(n-s));
	}
}