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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Scanner;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author xwchen
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskB solver = new TaskB();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskB {
        public void solve(int testNumber, Scanner in, PrintWriter out) {
            long n = in.nextLong();
            int res = 0;
            if (n < 9L) {
                res = (int) n;
            } else {
                res = gao(n);
            }

            out.println(res);

//        for(int i=9;i<=30000;++i)
//        {
//            //
//            if(fix(i)!=gao(i)){
//                out.println("FFFFFFFFF");
//                out.println(i+" -> "+ fix(i) +" --- "+ gao(i) );
//            }
//        }
        }

        int gao(long n) {
            int res = 0;
            long now = 9;
            while (now * 10 + 9 <= n) {
                now = now * 10 + 9;
            }
            res = Math.max(res, S(now) + S(n - now));
            return res;
        }

        int S(long x) {
            int sum;
            for (sum = 0; x > 0L; x /= 10L) {
                sum = (int) ((long) sum + x % 10L);
            }

            return sum;
        }

    }
}