#include <bits/stdc++.h>
using namespace std;

int main() {
	int n;
	cin>>n;
	int an=0;
	for(int i=1;i<=n;i++){
		int o=0,v=i;
		while(v){
			o^=1;
			v/=10;
		}
		an+=o;
	}
	cout<<an;
	return 0;
}
### Prompt

Construct a CPP code solution to the problem outlined:

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 main() {
	int n;
	cin>>n;
	int an=0;
	for(int i=1;i<=n;i++){
		int o=0,v=i;
		while(v){
			o^=1;
			v/=10;
		}
		an+=o;
	}
	cout<<an;
	return 0;
}
```