#include<bits/stdc++.h>
using namespace std;
int main() {
	int n;
	cin>>n;
	int ans=0;
	for(int i=1;i<=n;i*=100) {
		int mx=i*10-1;
		mx=min(mx, n);
		ans+=mx-i+1;
	}
	cout<<ans<<"\n";
	return 0;
}

### 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 main() {
	int n;
	cin>>n;
	int ans=0;
	for(int i=1;i<=n;i*=100) {
		int mx=i*10-1;
		mx=min(mx, n);
		ans+=mx-i+1;
	}
	cout<<ans<<"\n";
	return 0;
}

```