#include <iostream>
using namespace std;
int main(){
  int n, r = 0;
  cin >> n;
  for(int i = 1; i <= n; i++){
    if(to_string(i).length() % 2 == 1)r++;
  }
  cout << r << '\n';
}
### Prompt

Develop a solution in Cpp to the problem described below:

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, r = 0;
  cin >> n;
  for(int i = 1; i <= n; i++){
    if(to_string(i).length() % 2 == 1)r++;
  }
  cout << r << '\n';
}
```