#include "bits/stdc++.h"
using namespace std;
int main() {
	int N; cin >> N;
	int ans = 0;
	for (int i = 1; i <= N; i++) {
		int const log = log10(i) + 1;
		if (log % 2 == 1) ans++;
	}
	cout << ans << endl;
}
### Prompt

Please formulate a Cpp solution to 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 ans = 0;
	for (int i = 1; i <= N; i++) {
		int const log = log10(i) + 1;
		if (log % 2 == 1) ans++;
	}
	cout << ans << endl;
}
```