#include <bits/stdc++.h>
using namespace std;
int main() {
  int n,a=0; cin>>n;
  cout<<(n<10?n:n<100?9:n<1000?n-90:n<10000?909:n<100000?n-9090:90909);
}
### Prompt

Construct a Cpp code solution to the problem outlined:

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=0; cin>>n;
  cout<<(n<10?n:n<100?9:n<1000?n-90:n<10000?909:n<100000?n-9090:90909);
}
```