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

Your challenge is to write 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,m=0,a,b;
	scanf("%d",&n);
	for(int i=1;i<=n;i++){
		a=i;b=0;
		while(a>0){
			a/=10;
			b++;
		}
		if(b%2==1)	m++;
	}
	printf("%d",m);
	return 0;
}
```