#include <bits/stdc++.h>
using namespace std;
int main(){
  long n,a;
  cin >> n;
  long x = n,k=0,l=0;
  while(x>0){
      a = x%10;
      x /= 10;
      k++;
  }
  if((n+1)%long(pow(10,k-1))==0) cout << 9*(k-1)+a << endl;
  else cout << 9*(k-1)+a-1 << endl;
}
### Prompt

Generate 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 n,a;
  cin >> n;
  long x = n,k=0,l=0;
  while(x>0){
      a = x%10;
      x /= 10;
      k++;
  }
  if((n+1)%long(pow(10,k-1))==0) cout << 9*(k-1)+a << endl;
  else cout << 9*(k-1)+a-1 << endl;
}
```