#include <iostream>
#include <cmath>
using namespace std;

int main() {
  int a;
  cin >> a;

  int result = 0;
  for(int i=1; i <= a; i++) if((int)log10(i) % 2 == 0) result++;
  cout << result << endl;
}

### 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>
#include <cmath>
using namespace std;

int main() {
  int a;
  cin >> a;

  int result = 0;
  for(int i=1; i <= a; i++) if((int)log10(i) % 2 == 0) result++;
  cout << result << endl;
}

```