#include<bits/stdc++.h>
int main()
{
	int n,sum=0;
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	{
		int t=0,k=i;
		while(k!=0)
		{
			t=t+1;
			k=k/10;
		}
		if(t%2==1)sum++;
	}
	printf("%d\n",sum);
}
### 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>
int main()
{
	int n,sum=0;
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	{
		int t=0,k=i;
		while(k!=0)
		{
			t=t+1;
			k=k/10;
		}
		if(t%2==1)sum++;
	}
	printf("%d\n",sum);
}
```