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)

#define n_max 20
#define max(a,b) a>b?a:b
#include <stdio.h>
int main(void){
    char n[n_max];
    for(int i=0;i<n_max;i++)
    {
        n[i] = 0;
    }
    gets(n);
    int len = getlength(n);
    
    int nonine = len-1;
    for(int i=0;i<len;i++)
    {
        if(n[i]!='9')
        {
            nonine = i;
            break;
        }
    }
    
    if(nonine==len-1)
    {
        printf("%d\n",max(9*nonine+(n[nonine]-'0'),9*(nonine-1)+(n[nonine-1]-'1')+9));
        return 0;
    }
    printf("%d\n",9*nonine+n[nonine]-'1'+9*(len-nonine-1));
    return 0;
}

int getlength(char data[]){
    int i=0;
    while(data[i])
    {
        i++;
    }
    return i;
}