#include <bits/stdc++.h>
using namespace std;

int main() {
  long N;
  cin>>N;
  long ans =0;
  int flg=1;
  while(N>=10){
    ans+=9;
    if(N%10!=9) flg=0;
    N/=10;
  }
  if(flg)ans += N;
  else ans+=N-1;
  cout<<ans<<endl;
  }

### Prompt

Please provide a CPP coded solution to the problem described below:

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.

Examples

Input

100


Output

18


Input

9995


Output

35


Input

3141592653589793


Output

137

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;

int main() {
  long N;
  cin>>N;
  long ans =0;
  int flg=1;
  while(N>=10){
    ans+=9;
    if(N%10!=9) flg=0;
    N/=10;
  }
  if(flg)ans += N;
  else ans+=N-1;
  cout<<ans<<endl;
  }

```