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.io.*;
import java.util.*;
 
public class Solution {
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        long N = Long.parseLong(br.readLine());
        System.out.println(solve(N));
    }
    
    private static int solve(long N){
        if(N < 10) return (int)N;
        int sum = 0 , carry = 0;
        
        // all digits 9 check
        long temp = N;
        while(temp > 0){
            sum += temp % 10;
            if(temp % 10 != 9){
                sum = 0;
                break;
            }
            temp /= 10;
        }
        
        if(sum > 0) return sum;
        
        
        while(N > 0){
            int d = (int)(N % 10);
            N /= 10;
            if(N == 0){
                sum += d - carry;
            }else{
                d = d + 10 > 18 && carry == 0 ? 9 : d + 10;
                int d2 = (d - 9 - carry);
                sum += 9 + d2;
                carry = d / 10;
            }
        }
        
        return sum;
    }
}