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

int main() {
  long long N,keta=0;
  cin >> N;
  N++;
  long long n = N;
  for(int i=0; n>=1; i++){
    if(n/10 == 0){
      keta++;
      break;
    }
    n /= 10;
    keta++;
  }
  cout << (keta-1)*9+n-1 << endl;
}

### Prompt

Develop a solution in CPP 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 long N,keta=0;
  cin >> N;
  N++;
  long long n = N;
  for(int i=0; n>=1; i++){
    if(n/10 == 0){
      keta++;
      break;
    }
    n /= 10;
    keta++;
  }
  cout << (keta-1)*9+n-1 << endl;
}

```