#include <iostream>
using namespace std;

int main(){
	int n;
	cin >> n;
	int odd = 0;
	odd = min(n, 9);
	odd += min(max(n - 99, 0), 900);
	odd += min(max(n - 9999, 0), 90000);
	cout << odd << endl;
	return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 <iostream>
using namespace std;

int main(){
	int n;
	cin >> n;
	int odd = 0;
	odd = min(n, 9);
	odd += min(max(n - 99, 0), 900);
	odd += min(max(n - 9999, 0), 90000);
	cout << odd << endl;
	return 0;
}

```