#include<bits/stdc++.h>
using namespace std;
int main()
{
	int n,ans=0,i=1;
	cin>>n;
	while(i<n+1)
	{
		int size = trunc(log10(i)) + 1;
		if(size%2==1)
		++ans;
		++i;
	}
	cout<<ans;
}
### Prompt

Generate 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+1)
	{
		int size = trunc(log10(i)) + 1;
		if(size%2==1)
		++ans;
		++i;
	}
	cout<<ans;
}
```