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"bits/stdc++.h"
using namespace std;

#define rep(i,n) for(int (i)=0;(i)<(n);(i)++)
#define rep3(i,m,n) for(int (i)=m;(i)<=(n);(i)++)
#define rep3rev(i,m,n) for(int (i)=m;(i)>=(n);(i)--)
#define all(a) (a.begin()),(a.end())
#define rall(a) (a.rbegin()),(a.rend())
#define fi first
#define se second
#define pb push_back
#define eb emplace_back

using ll = long long;
using vll = vector<ll>;
using vi = vector<int>;
using vvi = vector<vector<int>>;
using P = pair<int, int>;
using Pl = pair<ll, int>;

void Main() {
    string s;
	cin >> s;
    int n = s.size();
    bool allnine = true;
    
    rep(i, n){
        if(i == 0) continue;
        if(s[i] - '0' != 9) {
            allnine = false;
            break;
        }
    }
    
    int ans;
    if(allnine){
        ans = s[0] - '0';
        ans += 9*(n-1);
    }else{
        ans = s[0] - '0' - 1;
        ans += 9*(n-1);
    }
    
    cout << ans << endl;
    return;
}

int main() {
	cin.tie(nullptr);
	ios_base::sync_with_stdio(false);
	cout << fixed << setprecision(15);
	Main();
	return 0;
}