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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <vector>

using namespace std;

int main()
{
	long long n;
	cin >> n;
	
	vector<int> x;
	long long tmp = n;
	while(tmp != 0)
	{
		x.insert(x.begin(), tmp % 10);
		tmp /= 10;
	}
	
	int ans = 0;
	for(int i = 1; i < x.size(); i++)
	{
		if(x[i] == 9)
		{
			ans += x[i - 1];
			if(i == x.size() - 1)
			{
				ans += 9;
			}
		}
		else
		{
			ans += x[i - 1] - 1;
			ans += (x.size() - i) * 9;
			break;
		}
	}
	
	if(x.size() == 1)
	{
		ans = x[0];
	}
	
	cout << ans << endl;
}