#include "bits/stdc++.h"

using namespace std;


int main()
{

  int N;
  cin >> N;
  int cnt = 0;
  for (int i = 1; i <= N; i++) {
    cnt += to_string(i).length() % 2;
  }
  cout << cnt;
  return 0;
}


### Prompt

Please create a solution in CPP 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++) {
    cnt += to_string(i).length() % 2;
  }
  cout << cnt;
  return 0;
}


```