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<iostream>
#include<string>
#include<vector>
#include<algorithm>
#include<bitset>
#include<set>
#include<map>
#include<stack>
#include<queue>
#include<iomanip>
using namespace std;

#define repr(i, a, b) for (int i = (int)(a); i < (int)(b); i++)
#define rep(i, n) repr(i, 0, n)
#define INF 2e9
#define MOD 1000000007
#define LINF (long long)4e18

using ll = long long;


int main(){
    string N; cin >> N;
    int ans = 0;
    if(N.size() == 1){
        ans = N[0] - '0';
    }
    else{
        if(N[0] == 9){
            rep(i,N.size()){
                if(N[i] != 9){
                    ans -= 1;
                    repr(j,i,N.size()) ans += 9;
                }
                ans += 9;
            }
        }
        else{
            int ok = true;
            repr(i,1,N.size()){
                if(N[i] != 9) ok = false;
            }
            if(ok){
                ans += N[0] - '0';
                repr(j,1,N.size()) ans += 9;
            }
            else{
                ans += N[0] - '0' - 1;
                repr(j,1,N.size()) ans += 9;
            }
        }
    }
    cout << ans << endl;
}