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

int main(){
  int n,ans=0;
  cin>>n;
  for(int i=1; i<=n; i++){
    int m=log10(i)+1;
    if(m%2==1) ans++;
  }
  cout<<ans<<endl;
}
### Prompt

Generate 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,ans=0;
  cin>>n;
  for(int i=1; i<=n; i++){
    int m=log10(i)+1;
    if(m%2==1) ans++;
  }
  cout<<ans<<endl;
}
```