#include <bits/stdc++.h>
using namespace std;
int main() {
  int n,a,b=0;
  cin >> n;
  for(int i=1;i<n+1;i++) {
    string s=to_string(i);
    a=s.length();
    b=b+a%2;
  }
  cout << b << 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,a,b=0;
  cin >> n;
  for(int i=1;i<n+1;i++) {
    string s=to_string(i);
    a=s.length();
    b=b+a%2;
  }
  cout << b << endl;
  return 0;
}

```