#include<iostream>
using namespace std;
int main(){
	int n,k=1,ans=0,i=10;
	cin>>n;
	if(n<10){
		cout<<n;
		return 0;
	}
	for(;i<=n;i*=100){
		ans+=i*0.9;
	}
	if(n/(i/100)>9)ans+=n-i/10+1;
	cout<<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<iostream>
using namespace std;
int main(){
	int n,k=1,ans=0,i=10;
	cin>>n;
	if(n<10){
		cout<<n;
		return 0;
	}
	for(;i<=n;i*=100){
		ans+=i*0.9;
	}
	if(n/(i/100)>9)ans+=n-i/10+1;
	cout<<ans;
}

```