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

int digitsum(string s) {
  int ans = 0;
  for ( char d : s) ans += d-'0';
  return ans;
}

int main() {
  string s;
  cin>>s;
  int n = s.length();
  cout << max(digitsum(s),s[0]-'1'+9*(n-1)) << endl;

  return 0;
}
### 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;

int digitsum(string s) {
  int ans = 0;
  for ( char d : s) ans += d-'0';
  return ans;
}

int main() {
  string s;
  cin>>s;
  int n = s.length();
  cout << max(digitsum(s),s[0]-'1'+9*(n-1)) << endl;

  return 0;
}
```