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.BufferedReader;
import java.io.File;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintStream;
import java.io.PrintWriter;
import java.lang.reflect.Array;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.HashSet;
import java.util.List;
import java.util.Map;
import java.util.Set;
import java.util.StringTokenizer;

/**
 * Created by timur on 28.03.15.
 */

public class TaskB {
    boolean eof;
    BufferedReader br;
    StringTokenizer st;
    PrintWriter out;

    public static void main(String[] args) throws IOException {
        new TaskB().run();
    }

    public String nextToken() {
        while (st == null || !st.hasMoreTokens()) {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (Exception e) {
                eof = true;
                return "-1";
            }
        }
        return st.nextToken();
    }

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

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

    double nextDouble() {
        return Double.parseDouble(nextToken());
    }

    String nextLine() throws IOException {
        return br.readLine();
    }


    void run() throws IOException {
        InputStream input = System.in;
        PrintStream output = System.out;
        try {
            File f = new File("a.in");
            if (f.exists() && f.canRead()) {
                input = new FileInputStream(f);
                output = new PrintStream("a.out");
            }
        } catch (Throwable e) {
        }
        br = new BufferedReader(new InputStreamReader(input));
        out = new PrintWriter(output);
        solve();
        br.close();
        out.close();
    }

    long gcd(long a, long b) {
        if (b == 0)
            return a;
        else
            return gcd(b, a % b);
    }


    void solve() {
        long n = nextLong();
        long x = n;
        long s = 0;
        long ans = 0;
        if (n < 10) {
            out.print(n);
            return;
        }
        while (n > 9) {
            n /= 10;
        }
        s = n - 1;
        ans += s;
        n = x;
        while (s * 10 + 9 <= n) {
            s = s * 10 + 9;
            ans += 9;
        }
        n = n - s;
        while (n > 0) {
            ans += n % 10;
            n /= 10;
        }
        out.print(ans);
    }
}