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: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdbool.h>
#include <math.h>

int main(void){
    long long n;
    scanf("%lld",&n);
    long long n1=n;
    int count=0,a[20],ans;
    bool nine=false;
    while (n1!=0){
        a[count]=n1%10;
        n1/=10;
        count++;
    }
    ans=a[count-1];
    for (int i=1; i<count; i++){
        if (nine==true){
            ans+=9;
        } else if (a[count-1-i]==9){
            ans+=9;
        } else {
            ans+=8;
            nine=true;
        }
    }
    printf("%d\n",ans);
	return 0 ;
}