#include<bits/stdc++.h>
using namespace std;
int main()
{
	int N,i,o=0;
	cin>>N;
	for(i=1;i<=N;i++)
	{
		int j=0;
		int p=i;
		while(p)
		{
			p/=10;
			j++;
		}
		if(j&1)
			o++;
	}
	cout<<o<<endl;
}
### Prompt

Develop a solution in Cpp to the problem described below:

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,i,o=0;
	cin>>N;
	for(i=1;i<=N;i++)
	{
		int j=0;
		int p=i;
		while(p)
		{
			p/=10;
			j++;
		}
		if(j&1)
			o++;
	}
	cout<<o<<endl;
}
```