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

int main() {
  int N, cnt=0;
  cin >> N;
  for(int i=1; i<=N; i++){
    string S=to_string(i);
    if(S.size()%2==1) cnt++;
  }
  cout << cnt << endl;
}
### Prompt

In cpp, your task is to solve 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, cnt=0;
  cin >> N;
  for(int i=1; i<=N; i++){
    string S=to_string(i);
    if(S.size()%2==1) cnt++;
  }
  cout << cnt << endl;
}
```