#include <bits/stdc++.h>

using namespace std;

int main() {
	long long N;
	cin >> N;
	long long length = log10(N);
	long long sum = 9 * length;
	N -= pow(10, length) - 1;
	for(int i = 0; i < length; i++)
		N /= 10;
	sum += N;
	cout << sum << endl;
}
### Prompt

Please provide a Cpp coded solution 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;
	cin >> N;
	long long length = log10(N);
	long long sum = 9 * length;
	N -= pow(10, length) - 1;
	for(int i = 0; i < length; i++)
		N /= 10;
	sum += N;
	cout << sum << endl;
}
```