#include<bits/stdc++.h>
using namespace std;
int d(int n){
	if(n==0)return 0;
  	return 1+d(n/10);
}
int main(){
  	int n,ans=0;cin>>n;for(int i=1;i<=n;i++){ans+=(d(i)&1);}
  	cout<<ans;
}
### Prompt

Develop a solution in CPP to the problem described below:

Given is an integer N. Find the number of positive integers less than or equal to N that have an odd number of digits (in base ten without leading zeros).

Constraints

* 1 \leq N \leq 10^5

Input

Input is given from Standard Input in the following format:


N


Output

Print the number of positive integers less than or equal to N that have an odd number of digits.

Examples

Input

11


Output

9


Input

136


Output

46


Input

100000


Output

90909

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;
int d(int n){
	if(n==0)return 0;
  	return 1+d(n/10);
}
int main(){
  	int n,ans=0;cin>>n;for(int i=1;i<=n;i++){ans+=(d(i)&1);}
  	cout<<ans;
}
```