#include <bits/stdc++.h>
using namespace std;
int main()
{
	int n;
	cin>>n;
	int cnt=0;
	for(int i=1;i<=n;i++)
		if ((i>=1&&i<=9)||(i>=100&&i<=999)||(i>=10000&&i<=99999))
			cnt++;
	cout<<cnt;
	return 0;
}
### Prompt

Please formulate a cpp solution to 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()
{
	int n;
	cin>>n;
	int cnt=0;
	for(int i=1;i<=n;i++)
		if ((i>=1&&i<=9)||(i>=100&&i<=999)||(i>=10000&&i<=99999))
			cnt++;
	cout<<cnt;
	return 0;
}
```