#include <bits/stdc++.h>
using namespace std;
long long n;
long long ans=0;
int main()
{
	scanf("%lld",&n);
	if (n<=9) 
	{
		printf("%lld\n",n);
		return 0;
	}
	n+=1;
	while (n>=10)
	{
		ans+=9;
		n/=10;
	}
	ans+=n-1;
	printf("%lld\n",ans);
	return 0;
}

### Prompt

In cpp, your task is to solve 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;
long long n;
long long ans=0;
int main()
{
	scanf("%lld",&n);
	if (n<=9) 
	{
		printf("%lld\n",n);
		return 0;
	}
	n+=1;
	while (n>=10)
	{
		ans+=9;
		n/=10;
	}
	ans+=n-1;
	printf("%lld\n",ans);
	return 0;
}

```