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>
#include <stdlib.h>
#include <math.h>
int main(void){
	int ans,keta=0,c,i;
	long long N,n;
	scanf("%lld",&N);
	n=N;
	while(n>=1){n/=10;keta++;}
	c = N/(pow(10,keta-1));
	ans = c+9*(keta-1);
	n=0;
	int tmp=1;
	for(i=0;i<keta;i++){
		if(i==keta-1) n+= c*tmp;
		else n += 9*tmp;
		tmp*10;
	}
	if(n!=N)ans--;
	
	printf("%d",ans);
	return 0;
}