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: Wrong Answer)

#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <iostream>
#include <climits>
#include <cstring>
#include <functional>

using namespace std;

using ll = long long;

#define rep(i, N) for(int i = 0;i < N;i++)
#define LARGE INT_MAX / 2

string N;

int main() {
	cin >> N;
	int ans = 0;
	rep(i, N.length()) {
		int num = 0;
		num = N[i] - '0';
		if (num == 9) {
			ans += 9;
		}
		else if(N[i + 1] != '9' && i != N.length() - 1){
			ans += num - 1 + 9 * (N.length() - i - 1);
			break;
		}
		else {
			ans += num;
		}
	}
	cout << ans << endl;
	return 0;
}