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

int main() {
	int l,cnt=0;
	cin >> l;
	
	for(int i=1;i<=l;i++) {
		int k = log10(i)+1;
		if(k % 2!=0) {cnt++;}
		else continue;
	}
	cout << cnt << endl;
	return 0;
}
### 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;

int main() {
	int l,cnt=0;
	cin >> l;
	
	for(int i=1;i<=l;i++) {
		int k = log10(i)+1;
		if(k % 2!=0) {cnt++;}
		else continue;
	}
	cout << cnt << endl;
	return 0;
}
```