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

int main()
{
	int n,ans=0;
	scanf("%d",&n);
	ans+=min(9,n);
	ans+=min(900,max(0,n-99));
	ans+=min(90000,max(0,n-9999));
	printf("%d\n",ans);
}

### 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 main()
{
	int n,ans=0;
	scanf("%d",&n);
	ans+=min(9,n);
	ans+=min(900,max(0,n-99));
	ans+=min(90000,max(0,n-9999));
	printf("%d\n",ans);
}

```