#include <bits/stdc++.h>
using namespace std;


int main() {
	// your code goes here
	long n,cnt=0;
	cin>>n;
	for(long i=1;i<=n;++i)
	{
	    string a=to_string(i);
	    if(a.size() & 1)cnt++;
	}
	cout<<cnt;
}

### Prompt

Create a solution in Cpp for 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 main() {
	// your code goes here
	long n,cnt=0;
	cin>>n;
	for(long i=1;i<=n;++i)
	{
	    string a=to_string(i);
	    if(a.size() & 1)cnt++;
	}
	cout<<cnt;
}

```