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

int main(){
	int n;
	cin >> n;
	int cnt = 0;
	for(int i=n;i>=1;i--) {
		if(to_string(i).size() % 2 == 1) cnt++;
	}
	cout << cnt << endl;
}
### Prompt

In Cpp, your task is to solve 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;
using ll = long long;

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