#include <bits/stdc++.h>
using namespace std;
int main() {
	string s;
	cin >> s;
	int sum = 0;
	for (int i = 0; i < (int)s.size(); i++) {
		sum += s[i] - '0';
	}
	int s2 = (s[0] - '0') + 9 * ((int)s.size() - 1);
	if (sum < s2) s2--;
	cout << s2 << endl;
	return 0;
}
### 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() {
	string s;
	cin >> s;
	int sum = 0;
	for (int i = 0; i < (int)s.size(); i++) {
		sum += s[i] - '0';
	}
	int s2 = (s[0] - '0') + 9 * ((int)s.size() - 1);
	if (sum < s2) s2--;
	cout << s2 << endl;
	return 0;
}
```