#include <bits/stdc++.h>
using namespace std;
int main() {
	int n;
	cin >> n;
	int ans = 0;
	for (int i = n; i > 0; i--) {
		ans += (int(log10(i) + 1) & 1);
	}
	cout << ans;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 ans = 0;
	for (int i = n; i > 0; i--) {
		ans += (int(log10(i) + 1) & 1);
	}
	cout << ans;
}

```