#include<bits/stdc++.h>
using namespace std;
int n,a=1,b=10,ans;
int main(){
	scanf("%d",&n);
	while(a<=n){
		for(int i=a;i<=min(b-1,n);i++)ans++;
		a*=100;b*=100;
	}
	printf("%d",ans);
}
### Prompt

In cpp, your task is to solve 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<bits/stdc++.h>
using namespace std;
int n,a=1,b=10,ans;
int main(){
	scanf("%d",&n);
	while(a<=n){
		for(int i=a;i<=min(b-1,n);i++)ans++;
		a*=100;b*=100;
	}
	printf("%d",ans);
}
```