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 <bits/stdc++.h>

using namespace std;

typedef long long LL;
typedef vector<int> VI;
typedef pair<int, int> PI;

int main() {
	string n;
	cin >> n;
	map<int, int> mp;
	for (int i = 0; i < n.size(); i++) {
		mp[n[i] - '0']++;
	}
	int ans = 0;
	if (mp.size() == 2 && !mp.count(0)) {
		ans = 0;
		for (auto e : n) {
			ans += (e - '0');
		}
		cout << ans << endl;
		return 0;
	}
	ans = (n.size() - 1) * 9;
	cout << ans + n[0] - '0' - 1 << endl;
	return 0;
}