#include <stdio.h>
#include <string>
int main(){
  int n,ans=0;
  scanf("%d",&n);
  for(int i=1;i<=n;i++){
    if(0!=std::to_string(i).length()%2){
      ans++;
    }
  }
  printf("%d",ans);
}
### 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 <stdio.h>
#include <string>
int main(){
  int n,ans=0;
  scanf("%d",&n);
  for(int i=1;i<=n;i++){
    if(0!=std::to_string(i).length()%2){
      ans++;
    }
  }
  printf("%d",ans);
}
```