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 <fstream>
#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <string>
#include <sstream>
#include <map>
#include <set>
#include <vector>
#include <stack>
#include <cmath>
#include <queue>
#include <random>

using namespace std;

#define INT_MAX_VALUE 2147483647
#define LONG_LONG_MAX_VALUE 9223372036854775807
#define ll long long

template <class T>
T mymax(T a,T b){
    if(a>=b) return a;
    return b;
}
template <class T>
T mymin(T a,T b){
    if(a<=b) return a;
    return b;
}

ll gcd(ll a, ll b){
    if(a<b){
        swap(a,b);
    }
    while(b){
        ll r = a%b;
        a=b;
        b=r;
    }
    return a;
}

ll lcm(ll a, ll b){
    return (a*b)/gcd(a,b);
}

struct XX{
    ll a;
    int b;
    int j;
    int i;
};

class xxGreater {
public:
    bool operator()(const XX& riLeft, const XX& riRight) const {
        //第2条件
        if((riLeft.a) == (riRight.a)){
            return riLeft.i < riRight.i;//<:昇順(小さいものから順番)、>:降順(大きいものから順番)
            //プライオリティキューの場合は > で、top()すると値の小さいものがとれる
        }
        //第1条件
        return (riLeft.a) < (riRight.a);
    }
};


//map<long long,long long> prime_f(long long n){
//    map<long long,long long>res;
//    for(int i=2;i*i<=n;i++){
//        while(n%i==0){
//            ++res[i];
//            n/=i;
//        }
//    }
//    if(n!=1)res[n]=1;
//    return res;
//}


int main(int argc, const char * argv[])
{
    //scanf("%s",S);
    //scanf("%d",&N);
    //scanf("%lld %lld",&target1,&target2);
    //sscanf(tmp.c_str(),"%dd%d%d",&time[i], &dice[i], &z[i]);
    //getline(cin, target);
    //ifstream ifs( "1_06.txt" );//テスト用
    //ifs >> a;
    //ここから
    
    //入力高速化
    ios::sync_with_stdio(false);
    cin.tie(0);
    
    string s;
    cin >> s;
    if(s.size()==1){
        cout << s << endl;
        return 0;
    }
    int ans=0;
    for(int i=0;i<s.size()-1;i++){
        ans+=9;
    }
    ans+=(*s.substr(0,1).c_str()-'0')-1;
    
    int ans2=0;
    for(int i=0;i<s.size()-1;i++){
        ans2+=(*s.substr(i,1).c_str()-'0');
    }
    cout << max(ans,ans2) << endl;

    //ここまで
    //cout << "ans" << endl;改行含む
    //printf("%.0f\n",ans);//小数点以下表示なし
    //printf("%.7f\n",p);
    
    return 0;
}