#include<iostream>
#include<cmath>
using namespace std;

int main(){
	int n;
	cin >> n;
	int count =0;
	for (size_t i = 1; i <= n; i++)
	{	
		if(((int)log10(i)) %2 == 0) count++;
	}

	cout << count;
	
}
### 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<iostream>
#include<cmath>
using namespace std;

int main(){
	int n;
	cin >> n;
	int count =0;
	for (size_t i = 1; i <= n; i++)
	{	
		if(((int)log10(i)) %2 == 0) count++;
	}

	cout << count;
	
}
```