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 GetDigit(double num){
	return log10(num) + 1;
}

int main(void)
{	
	double a;
	

	int sum = 0;
	int max = 0;
	int num,i_copy;
	if (scanf("%lf", &a)==1);
	int num_1 = GetDigit(a);

	if (num_1 == 1);

	for (int i = ((int)a / 10 * (num_1 - 1)) * 10 * (num_1 - 1); i <= a; i++){
		num = GetDigit(i);
		i_copy = i;
		for (int j = 0; j < num; j++){
			sum += i_copy % 10;
			i_copy /= 10;
		}
		
		if (sum > max){
			max = sum;
		}
		sum = 0;
	}
	
	printf("%d\n",max);

	return 0;

}