#include <cstdio>
int main()
{
	int n , i , ans = 0;
	scanf("%d" , &n);
	for(i = 10 ; i <= n ; i *= 100)
		ans += i - i/10;
	i /= 10;
	if(n >= i) ans += n - i + 1;
	printf("%d\n" , ans);
	return 0; 
}
### 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 <cstdio>
int main()
{
	int n , i , ans = 0;
	scanf("%d" , &n);
	for(i = 10 ; i <= n ; i *= 100)
		ans += i - i/10;
	i /= 10;
	if(n >= i) ans += n - i + 1;
	printf("%d\n" , ans);
	return 0; 
}
```