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

Please provide a Cpp coded solution 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>
#include<cmath>
int main(){
	int n,ans=0;
	scanf("%d",&n);
	for(int i=1;i<=n;i++){
	int sum=0;
	int j=i;
		while(j>0){
			j/=10;
			sum++;
		}
	if(sum%2==1)
	ans++;
	}
	printf("%d",ans);
}
```