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.PrintWriter;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;
import java.util.*;

public class B {

    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);

        Solver solver = new Solver();
        
        solver.solve(1, in, out);

        out.close();
    }

    static class Solver {

        public void solve(int testNumber, InputReader in, PrintWriter out) {            
            long n = in.nextLong();
            if (n < 10) {
                out.println(n);
            } else {
                int len = Long.toString(n).length();
                long[] digits = new long[len];
                long k = n;
                int cur = 0;
                while (k > 0) {
                    digits[cur++] = k % 10;
                    k /= 10;
                }
                k = n;
                long pwr = 1L;
                for (int i = 0 ; i < len - 1 ; i++) {
                    k-= digits[i] * pwr;
                    pwr *= 10L;
                }
                k -= 1L;
                out.println(F(k) + F(n - k));
            }
        }

        public long F(long n) {
            long res = 0;
            while (n > 0) {
                res += n % 10;
                n /= 10;
            }
            return res;
        }
    }

    static class InputReader {

        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

        public long nextLong() {
            return Long.parseLong(next());
        }

        public double nextDouble() {
            return Double.parseDouble(next());
        }

        public float nextFloat() {
            return Float.parseFloat(next());
        }
    }
}