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

int main()
{
    int n;
    cin >> n;
    int cnt = 0;

    for (int i = 1; i <= n; i++) {
        if ((int)log10(i)%2 == 0) cnt++;
    }

    cout << cnt << "\n";
}

### 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 main()
{
    int n;
    cin >> n;
    int cnt = 0;

    for (int i = 1; i <= n; i++) {
        if ((int)log10(i)%2 == 0) cnt++;
    }

    cout << cnt << "\n";
}

```