#include <bits/stdc++.h>
using namespace std;
int main(){
  long long N; cin>>N;
  N+=1;
  long long a=log10(N);
  long long b=pow(10,a);
  if(N/b==1)cout<<a*9<<endl;
  else cout<<a*9+N/b-1<<endl;
}
### Prompt

Please formulate a Cpp solution to the following problem:

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 long N; cin>>N;
  N+=1;
  long long a=log10(N);
  long long b=pow(10,a);
  if(N/b==1)cout<<a*9<<endl;
  else cout<<a*9+N/b-1<<endl;
}
```