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.* ;
import java.io.BufferedReader ;
import java.io.InputStreamReader ;

public class MaximumSumOfDigits
{
    private static final boolean debug = false ;
    static int getDigits(long num)
    {
        int n = 1 ;
        if( num>=((int)1e8)) { n+=8 ; num/=(int)1e8 ;}
        if( num>=(int)1e4) { n+=4 ; num /= (int)1e4 ;} 
        if( num>= (int)1e2) { n+=2 ; num /= (int)1e2;}
        if( num>10) {n+=1;}
        return n ;
    }
    public static void main(String args[]) throws Exception 
    {
        BufferedReader bro = new BufferedReader(new InputStreamReader(System.in)) ;
        long n  = Long.parseLong(bro.readLine()) ;
        int dig = getDigits(n) ;
        long num1 = (n/((long)Math.pow(10,dig-1)))*(long)Math.pow(10,dig-1)-1 ;
        long num2 = n-num1 ;
        if(debug) 
        {
            System.out.println("num1 :"+num1 + " num2 :"+num2+" dig :"+dig) ;
        }
        System.out.println(sumD(num1)+sumD(num2)) ;
    }
    static int sumD(long n)
    {
        int sum = 0 ;
        while(n!=0)
        {
            sum+=n%10 ;
            n/=10 ;
        }
        return sum ;
    }
}