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

Your challenge is to write a Cpp solution to the following problem:

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 n,ans;
int main(){
	cin>>n;
	for(int i=1;i<=n;++i){
		int s=0,k=i;
		while(k){
			s++;k/=10;
		}
		if(s&1)ans++;
	}
	cout<<ans;
}
```