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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

using namespace std;
#define rep(i, n) for (int i = 0; i < n; i++)
#define ll long long int
#define INF 1000000007
#define llINF 1000000000000000007
#define v(i) vector<i>
#define vv(i) vector<vector<i>>
// cin.eof() 入力個数の問題
// int num = atoi(string.c_str()); 文字列の整数変換
int keta(ll n){
    int ans=0;
    while(n>0){
        n /= 10;
        ans++;
    }
    return ans;
}


int main() {
ll n;
cin>>n;
ll ans = 0;
ll cnt = 0,num = 9;
rep(i,keta(n)-1){
    cnt += num;
    num *= 10;
}
//cout<<cnt<<endl;
  ll as = 1;
  rep(i,keta(n)-1){
  as *= 10;
  }
for(int i = 9;i >= 0;i--){
    ll Q = i*as + cnt;
    if(Q <= n){
        cout<<i+(keta(n)-1)*9<<endl;
return 0;
    }
}


return 0;
}