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

int64_t dp(int64_t N){
  if(N < 10) return N;
  else       return max(dp(N/10)+ N%10, dp(N/10- 1)+ 9);
}

int main(){
  int64_t N;
  cin >> N;

  cout << dp(N) << endl;
}
### Prompt

Create a solution in Cpp for 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;

int64_t dp(int64_t N){
  if(N < 10) return N;
  else       return max(dp(N/10)+ N%10, dp(N/10- 1)+ 9);
}

int main(){
  int64_t N;
  cin >> N;

  cout << dp(N) << endl;
}
```