#include<bits/stdc++.h>
using namespace std;
int main() {
	int N,ans=0;
	cin >> N;
	
	for (int i = 1; i <= N; i++)
	{
		int l = to_string(i).length();
		if( l % 2 == 1) ++ans;
	}

	cout << ans << endl;
}
### 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,ans=0;
	cin >> N;
	
	for (int i = 1; i <= N; i++)
	{
		int l = to_string(i).length();
		if( l % 2 == 1) ++ans;
	}

	cout << ans << endl;
}
```