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

int main(void){
    int n,ans=0 ;
    cin >> n ;
    ans+=min(n,9) ;
    ans+=max(0,min(n-99,900)) ;
    ans+=max(0,min(n-9999,90000));
    cout << ans << endl ;
}
### Prompt

Your challenge is to write 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(void){
    int n,ans=0 ;
    cin >> n ;
    ans+=min(n,9) ;
    ans+=max(0,min(n-99,900)) ;
    ans+=max(0,min(n-9999,90000));
    cout << ans << endl ;
}
```