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

Construct a Cpp code solution to the problem outlined:

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 = 0, f = 1;
int main(){
	scanf("%d", &n);
	int x = 1;
	for(int i = 1; i <= n; i++){
		if(i / x >= 10) x *= 10, f = f ^ 1;
		ans += f;
	}
	printf("%d", ans);
	return 0;
}
```