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

int digit (long long int v) {
    int cnt = 1;
    while((v/=10) != 0) {
	cnt++;
    }
    return cnt;
}
int main() {
	
	long long int n,k,l;
	cin>>n;
	k=digit(n);
	l=pow(10,k-1);
	
	cout<<9*(k-1)+(n+1)/l-1<<endl;
}
### Prompt

Construct a Cpp code solution to the problem outlined:

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 digit (long long int v) {
    int cnt = 1;
    while((v/=10) != 0) {
	cnt++;
    }
    return cnt;
}
int main() {
	
	long long int n,k,l;
	cin>>n;
	k=digit(n);
	l=pow(10,k-1);
	
	cout<<9*(k-1)+(n+1)/l-1<<endl;
}
```