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;

/**
 * Contest: Codeforces Round #513, Problem B
 *
 * @author Arturs Licis
 *
 * @see <a href="http://codeforces.com/contest/1060/problem/B">Maximum Sum of Digits</a>
 */
public class ProblemB {
    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);

        long n = scan.nextLong();
        int sum = 0;

        char[] numChars = Long.toString(n).toCharArray();
        char char0 = (char) (numChars[0] - 1);
        sum += char0 - '0';
        numChars[0] = char0;
        for (int i = 1; i < numChars.length; i++) {
            numChars[i] = '9';
            sum += 9;
        }

        long num1 = Long.valueOf(String.valueOf(numChars));
        long num2 = n - num1;
        char[] numChars2 = Long.toString(num2).toCharArray();

        for (int i = 0; i < numChars2.length; i++) {
            sum += numChars2[i] - '0';
        }

        System.out.println(sum);
    }
}