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<algorithm>
#include<cstdio>
#include<cmath>
#include<cctype>
#include<math.h>
#include<string>
#include<string.h>
#include<stack>
#include<queue>
#include<vector>
#include<utility>
#include<set>
#include<map>
#include<stdlib.h>
#include<iomanip>

using namespace std;

#define ll long long
#define ld long double
#define EPS 0.0000000001
#define INF 1e9
#define MOD 1000000007
#define rep(i,n) for(i=0;i<(n);i++)
#define loop(i,a,n) for(i=a;i<(n);i++)
#define all(in) in.begin(),in.end()
#define shosu(x) fixed<<setprecision(x)

typedef vector<int> vi;
typedef vector<string> vs;
typedef pair<int,int> pii;

int main(void) {
  int i,j;
  string s;
  cin>>s;
  int n=s.size();
  n--;
  if(s[0]=='1')
    cout<<9*n<<endl;
  else 
    cout<<9*n+(s[0]-'1')<<endl;
}