#include <iostream>
using namespace std;

int n, c;

int main(){
	cin >> n;
	c = 0;
	for (int b=1; b<=n; b*=100){
		c += min(n - b + 1, 9*b);
	}
	cout << c;
}
### Prompt

Your challenge is to write 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 <iostream>
using namespace std;

int n, c;

int main(){
	cin >> n;
	c = 0;
	for (int b=1; b<=n; b*=100){
		c += min(n - b + 1, 9*b);
	}
	cout << c;
}
```