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 <cassert>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <bitset>
#include <iomanip>
#include <iostream>
#include <istream>
#include <map>
#include <numeric>
#include <ostream>
#include <queue>
#include <set>
#include <sstream>
#include <string>
#include <utility>
#include <vector>
using namespace std;
using ll = long long;
class Solution {
public:
    void solve(std::istream& in, std::ostream& out) {
        string n;
        in >> n;
        if (1 == n.size()) {
            out << n << endl;
        }
        else {
            bool check = true;
            for (int i = 0; i < n.size(); i++) {
                if (n[i] != '9') {
                    check = false;
                    break;
                }
            }
            if (!check) {
                for (int i = 1; i < n.size(); i++) {
                    n[i] = '0';
                }
            }
            ll num = atoll(n.c_str());
            if (!check) {
                num--;
            }
            ll ans = 0;
            while (0 < num) {
                ans += num % 10;
                num /= 10;
            }
            out << ans << endl;
        }
    }
};
void solve(std::istream& in, std::ostream& out) {
    out << std::setprecision(12);
    Solution solution;
    solution.solve(in, out);
}
#include <fstream>
int main() {
    
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    istream& in = cin;
    ostream& out = cout;
    solve(in, out);
    return 0;
}