#include <bits/stdc++.h>
using namespace std;
int n,ans=0;
int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;i++){
		int j=i,tmp=0;
		while(j) j/=10,tmp++;
		if(tmp&1) ans++;
	}
	printf("%d\n",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 <bits/stdc++.h>
using namespace std;
int n,ans=0;
int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;i++){
		int j=i,tmp=0;
		while(j) j/=10,tmp++;
		if(tmp&1) ans++;
	}
	printf("%d\n",ans);
} 
```