#include <bits/stdc++.h>
using namespace std;
int main(){
	int N,ans=0,i=1;
	cin >> N;
	while(i<=N){
		ans += min(i*10,N+1)-i;
		i *= 100;
	}
	cout << ans << endl;
}
### 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 main(){
	int N,ans=0,i=1;
	cin >> N;
	while(i<=N){
		ans += min(i*10,N+1)-i;
		i *= 100;
	}
	cout << ans << endl;
}
```