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.Arrays;
import java.util.Scanner;

public class t5 {
    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        long n = input.nextLong();
        long a = 0;
        long b = n;
        int length = String.valueOf(n).length();
        String str = "";
        for (int i = 0; i < length - 1; i++) {
            str += "9";
        }
        if (str.length() >= 1) {
            a = Long.parseLong(str);
            b = n - Long.parseLong(str);
        }else{
            if(n%2 == 0){
                a = n/2;
                b = n/2;
            }else{
                a = n/2;
                b = (n/2) +1;
            }
        }
        long s = sum(a , b);
        System.out.println(s);
        input.close();
    }

    public static long sum(long a, long b) {
        long s = 0;
        String s_a = String.valueOf(a);
        String s_b = String.valueOf(b);
        for (int i = 0; i < s_a.length(); i++) {
            s += Character.getNumericValue(s_a.charAt(i));
        }
        for (int i = 0; i < s_b.length(); i++) {
            s += Character.getNumericValue(s_b.charAt(i));
        }
        return s;
    }

}