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

long long N;

int main(int argc, char* argv[]){
  cin >> N;
  int l = log10(N);
  long long p = pow(10,l);
  cout << l*9+((N%p==p-1)?N/p:N/p-1) << endl;
  return 0;
}
### Prompt

Please create a solution in Cpp 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;

long long N;

int main(int argc, char* argv[]){
  cin >> N;
  int l = log10(N);
  long long p = pow(10,l);
  cout << l*9+((N%p==p-1)?N/p:N/p-1) << endl;
  return 0;
}
```