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.Scanner;
import java.util.Arrays;
public class Main {
	public static void main(String arg[]){
		Scanner sc = new Scanner(System.in);
		String n = sc.next();
		if(n.length() == 1) {
			System.out.println(n);
		}else{			
			int resp = 0;
			char[] x = new char[n.length()];
			long y;
			int i;
			int cont = 0;
			long div = 1;
			for(i=0; i<n.length()-1;i++) {
				cont += 9;
			}
			//System.out.println(cont);
			
			y = Long.parseLong(String.valueOf(n));
			for(i=0;i<n.length()-1; i++) {
				if(y*div%10 != 0 || y*div+div%10 != 0) {
					y -= 9*div;
					div = div*10;
				}
			}
			//System.out.println(y);
			resp += cont;
			String y1 = Long.toString(y);
			char[] vetor = y1.toCharArray();
			for(i=0; i<y1.length(); i++) {
				resp += Integer.parseInt(String.valueOf(vetor[i]));
			}
			System.out.println(resp);
		}	
	}
}