#include<bits/stdc++.h> 
using namespace std;
int main(){
  int N,sum=0;
  cin>>N;
  for(int i=1;i<=N;i++){
    string X=to_string(i);
    if(X.size()%2==1)
      sum++;
    }
  cout<<sum<<endl;
  return 0;
  }
### 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,sum=0;
  cin>>N;
  for(int i=1;i<=N;i++){
    string X=to_string(i);
    if(X.size()%2==1)
      sum++;
    }
  cout<<sum<<endl;
  return 0;
  }
```