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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
int main()
{
    char n[17];
    scanf("%s", n);
    int i, a, b, j;
    i = a = b = j = 0;
    int p;
    while (n[i] != '\0')
    {
        i++;
    }
    j = 0;
    for (b = 1; b < i; b++)
        if (n[b] != '9')
            j++;
    if (j == 0)
    {
        a = i * 9;
    }
    else
    {
        while (n[0] == '0')
        {
            for (j = 0; j < i - 1; j++)
                n[j] = n[j + 1];
            i--;
        }
        a = 9 * (i - 1) + (int)(n[0] - '0') - 1;
    }
    printf("%d\n", a);
    return 0;
}