#include <iostream>

using namespace std;

int main() {
    int N;
    cin >> N;
    int count = 0;
    for (int i = 1; i <= N; ++i) if (to_string(i).length() % 2 != 0) ++count;
    cout << count << endl;
}
### Prompt

Your challenge is to write 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 <iostream>

using namespace std;

int main() {
    int N;
    cin >> N;
    int count = 0;
    for (int i = 1; i <= N; ++i) if (to_string(i).length() % 2 != 0) ++count;
    cout << count << endl;
}
```