#include <bits/stdc++.h>
#include <math.h>
#include <string.h>
using namespace std;int f(long n){int p=log10(n);return 9*p+(int)(n/pow(10,p))-1;}int main(){long n;cin>>n;cout<<f(++n)<<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>
#include <math.h>
#include <string.h>
using namespace std;int f(long n){int p=log10(n);return 9*p+(int)(n/pow(10,p))-1;}int main(){long n;cin>>n;cout<<f(++n)<<endl;}
```