Problem:
Score : 300 points

Problem Statement

Find the maximum possible sum of the digits (in base 10) of a positive integer not greater than N.

Constraints

1\leq N \leq 10^{16}

N is an integer.

Input

Input is given from Standard Input in the following format:

N

Output

Print the maximum possible sum of the digits (in base 10) of a positive integer not greater than N.

Sample Input 1

100

Sample Output 1

18

For example, the sum of the digits in 99 is 18, which turns out to be the maximum value.

Sample Input 2

9995

Sample Output 2

35

For example, the sum of the digits in 9989 is 35, which turns out to be the maximum value.

Sample Input 3

3141592653589793

Sample Output 3

137

Solution in C# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
using System.Linq;
using System.Collections.Generic;
using static System.Console;

class Solver {
    Scanner sc = new Scanner();

    public void Solve() {
        var N = ReadLine();

        WriteLine((int.Parse(N[0].ToString()) - 1) + ((N.Length - 1) * 9));
    }
}



class Scanner {
    string[] s;
    int i;

    char[] cs = new char[] { ' ' };

    public Scanner() {
        s = new string[0];
        i = 0;
    }

    public string next() {
        if (i < s.Length) return s[i++];
        string st = Console.ReadLine();
        while (st == "") st = Console.ReadLine();
        s = st.Split(cs, StringSplitOptions.RemoveEmptyEntries);
        if (s.Length == 0) return next();
        i = 0;
        return s[i++];
    }

    public int nextInt() {
        return int.Parse(next());
    }
    public int[] ArrayInt(int N, int add = 0) {
        int[] Array = new int[N];
        for (int i = 0; i < N; i++) {
            Array[i] = nextInt() + add;
        }
        return Array;
    }

    public long nextLong() {
        return long.Parse(next());
    }

    public long[] ArrayLong(int N, long add = 0) {
        long[] Array = new long[N];
        for (int i = 0; i < N; i++) {
            Array[i] = nextLong() + add;
        }
        return Array;
    }

    public double nextDouble() {
        return double.Parse(next());
    }

    public double[] ArrayDouble(int N, double add = 0) {
        double[] Array = new double[N];
        for (int i = 0; i < N; i++) {
            Array[i] = nextDouble() + add;
        }
        return Array;
    }
}

class Program {
    static void Main(string[] args) {
        Solver s = new Solver();
        s.Solve();
    }
}