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

int main(){
	int N;
	cin >> N;
	int cnt = 0;
	for(int i = 1; i <= N; i++){
		string S = to_string(i);
		if(S.size() % 2 == 1) cnt++;
	}
	cout << cnt << endl;
}
### 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 <bits/stdc++.h>
using namespace std;

int main(){
	int N;
	cin >> N;
	int cnt = 0;
	for(int i = 1; i <= N; i++){
		string S = to_string(i);
		if(S.size() % 2 == 1) cnt++;
	}
	cout << cnt << endl;
}
```