#include<bits/stdc++.h>
using namespace std;
int n,s=0,t;
long long ans=0;
int main()
{
	cin>>n;
	for(int i=1;i<=n;i++){
		s=floor(log10(i))+1;
		if(s%2==1)ans++;
	}
	cout<<ans<<endl;
	return 0;
}

### Prompt

Develop a solution in cpp 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,s=0,t;
long long ans=0;
int main()
{
	cin>>n;
	for(int i=1;i<=n;i++){
		s=floor(log10(i))+1;
		if(s%2==1)ans++;
	}
	cout<<ans<<endl;
	return 0;
}

```