#include <iostream>

using namespace std;
int main(){
	int i,n,ans = 0;
	cin >> n;
	for(i=1;i<=n;i++){
		int x = i,cnt = 0;
		while(x){
			cnt++;
			x /= 10;
		}
		if(cnt&1) ans++;
	}
	cout << ans << endl;
}
### Prompt

Please provide a CPP coded solution to the problem described below:

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 i,n,ans = 0;
	cin >> n;
	for(i=1;i<=n;i++){
		int x = i,cnt = 0;
		while(x){
			cnt++;
			x /= 10;
		}
		if(cnt&1) ans++;
	}
	cout << ans << endl;
}
```