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(void)
{
	int a;
	int b[17];
	int i, j;
	int sum1 = 0, sum2 = 0;
	scanf("%d", &a);
	for(i=0; i<17; i++){
		b[i] = a % 10;
		a /= 10;
		if(a==0) break;
	}
	
	sum2 = b[i] - 1;
	for(j=0; j<=i; j++){
		sum1+=b[j];
		sum2+=9;
	}
	sum2 -= 9;

	if(sum1 > sum2) printf("%d\n", sum1);
	else printf("%d\n", sum2);

	return 0;
}