#include<bits/stdc++.h>
using namespace std;
int n, ans;

int main(){
	cin >> n;
	for(int i = 1; i <= n; i *= 100){
		if(i*10 <= n) ans += i*9;
		else ans += n - i + 1;
	}
	cout << ans << "\n";
}
### Prompt

Please formulate 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 *= 100){
		if(i*10 <= n) ans += i*9;
		else ans += n - i + 1;
	}
	cout << ans << "\n";
}
```