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)

/******************************************************************************

                              Online C++ Compiler.
               Code, Compile, Run and Debug C++ program online.
Write your code in this editor and press "Run" button to compile and execute it.

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#include <iostream>
#include <string>
#include <bits/stdc++.h>

using namespace std;

int main()
{
    string s;
    cin >>s;
    int sum=0;
  	if(s.length()>1){
      for(int i=0;i<s.length();i++){
          if(s[i]!='9'){
              if(i == (s.length()-1)){
                  if(stoi(s.substr(i-1,1))<8){
                  	sum += stoi(s.substr(i-1,1))-1;
                  }
               	  else{
                    sum += stoi(s.substr(i,1));
                  }
              }
              else{
                  if(i!=0){
                    if(stoi(s.substr(i-1,1))<8){
                      sum += stoi(s.substr(i-1,1))-1;
                    }
                    else{
                      sum += stoi(s.substr(i,1))-1;
                    } 
                  }
                  else{
                    sum += stoi(s.substr(i,1))-1;
                  }
              }
              break;
          }
      }
    
  	
      sum += 9*(s.length()-1);
      
    }
    else{
      sum += stoi(s.substr(0,1));
    }
    cout<<sum;
    return 0;
}