#include<cstdio>
int n,ans;
int main()
{
	scanf("%d",&n);
	for(int i=1,s;i<=n;i++)
	{
		s=0;
		for(int j=i;j;j/=10)s++;
		if(s%2)ans++;
	}
	printf("%d",ans);
	return 0;
}
### 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<cstdio>
int n,ans;
int main()
{
	scanf("%d",&n);
	for(int i=1,s;i<=n;i++)
	{
		s=0;
		for(int j=i;j;j/=10)s++;
		if(s%2)ans++;
	}
	printf("%d",ans);
	return 0;
}
```