#include<bits/stdc++.h>
using namespace std;
int main()
{
	int a,n=0,cnt,h;
	cin>>a;
	for(int i=0;i<=a;i++)
	{
		cnt=0;
		h=i;
		while(h>0)
		{
			cnt++;
			h=h/10;
		}
		if(cnt%2!=0){ n++;}
	}
	cout<<n;	
}
### 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 a,n=0,cnt,h;
	cin>>a;
	for(int i=0;i<=a;i++)
	{
		cnt=0;
		h=i;
		while(h>0)
		{
			cnt++;
			h=h/10;
		}
		if(cnt%2!=0){ n++;}
	}
	cout<<n;	
}
```