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

int main() {
  int N,count=0;
  cin >> N;
  
  for(int i=0;i<N;i++){
    int C=log10(i+1)+1;
    if(C%2==1){
      count++;
    }
  }
  cout << count << 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,count=0;
  cin >> N;
  
  for(int i=0;i<N;i++){
    int C=log10(i+1)+1;
    if(C%2==1){
      count++;
    }
  }
  cout << count << endl;
}     
```