#include <bits/stdc++.h>

using namespace std;

int n, ans;

int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; i++)
		ans += to_string(i).size() % 2;
	printf("%d\n", ans);
}

### Prompt

Generate 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 n, ans;

int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; i++)
		ans += to_string(i).size() % 2;
	printf("%d\n", ans);
}

```