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.math.BigInteger;
import java.util.Scanner;

public class q2B {

    private static Scanner input =  new Scanner(System.in);
    public static void main(String[] args) {

//        System.out.println(S(2197413));

//        double number = input.nextDouble();
//        double half = Math.floor(number / 2);
//        double remHalf = number - half;
//        int max = 0;
//        double i = 0;
//        while (true) {
//            int calc = S(half-i) + S(remHalf+i);
//            if (calc >= max)  max = calc;
//            else break;
//            i++;
//        }
//        System.out.println(max);

        long number = input.nextLong();
        long rem10 = number % 10;
//        System.out.println((S(9) + S(number-9)));
        int max  = 0;
        for (int i = 0; i < countS(number); i++) {
            double base  = Math.pow(10,i)-1;
            int calc =  S(base) + S(number - base);
            if (calc > max) max = calc;
        }
        System.out.println(max);





    }
    private static int S(double a)
    {
            int res = 0;
            while (a > 0)
            {
                res += a % 10;
                a = Math.floor(a/10);
            }
            return res;
    }
    private static int countS(double a)
    {
        int res = 0;
        while (a > 0)
        {
            res ++;
            a = Math.floor(a/10);
        }
        return res;
    }
}
//10000000000