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

int main(){
	int n,c=0;
	cin >> n;
	for(int i=1; i<=n; i++){
		if((to_string(i).length())%2!=0)c++;
	}
	cout << c <<endl;
	return 0;
}

### Prompt

Generate 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,c=0;
	cin >> n;
	for(int i=1; i<=n; i++){
		if((to_string(i).length())%2!=0)c++;
	}
	cout << c <<endl;
	return 0;
}

```