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

int main() {
  int N;
  cin >> N;
  int ans{};
  for (int i{1}; N >= i; i *= 100) {
    ans += min(i*10-1, N) - (i - 1);
  }

  cout << ans << endl;
  return 0;
}

### 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{};
  for (int i{1}; N >= i; i *= 100) {
    ans += min(i*10-1, N) - (i - 1);
  }

  cout << ans << endl;
  return 0;
}

```