#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll n;
int a[10010],tp,ans;
int main(){
	cin>>n;
	while(n)a[++tp]=n%10,n/=10;
	for(int i=tp,sum=0;i>=1;--i)if(a[i]){
		ans=max(ans,(i-1)*9+sum+a[i]-(i==1?0:1));
		sum+=a[i];
	}
	printf("%d",ans);
}

### 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;
typedef long long ll;
ll n;
int a[10010],tp,ans;
int main(){
	cin>>n;
	while(n)a[++tp]=n%10,n/=10;
	for(int i=tp,sum=0;i>=1;--i)if(a[i]){
		ans=max(ans,(i-1)*9+sum+a[i]-(i==1?0:1));
		sum+=a[i];
	}
	printf("%d",ans);
}

```