#include<cstdio>
using namespace std;
int main(){
	int n,ans=0;
	scanf("%d",&n);
	for (int i=1;i<=n;i++){
		int x=i,z=0;
		while (x) z++,x/=10;
		if (z%2) ans++;
	}
	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>
using namespace std;
int main(){
	int n,ans=0;
	scanf("%d",&n);
	for (int i=1;i<=n;i++){
		int x=i,z=0;
		while (x) z++,x/=10;
		if (z%2) ans++;
	}
	printf("%d\n",ans);
	return 0;
}
```