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


int main() {
	
	int N, cnt=0;
	cin >> N;
	for(int i=1; i<=N; i++){
		if(to_string(i).size()%2) cnt++;
		
	}
	cout << cnt << endl;
	
	return 0;
}
### Prompt

Create a solution in Cpp for 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, cnt=0;
	cin >> N;
	for(int i=1; i<=N; i++){
		if(to_string(i).size()%2) cnt++;
		
	}
	cout << cnt << endl;
	
	return 0;
}
```