#include<bits/stdc++.h>
using namespace std;
int n,cnt;
int main()
{
    cin>>n;
    for(int i=1;i<=n;i++){
        int x=(int(log10(i))+1);
        if(x%2==1)cnt++;
    }
    cout<<cnt<<endl;
    return 0;
}
### Prompt

Please provide a CPP coded solution to the problem described below:

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 n,cnt;
int main()
{
    cin>>n;
    for(int i=1;i<=n;i++){
        int x=(int(log10(i))+1);
        if(x%2==1)cnt++;
    }
    cout<<cnt<<endl;
    return 0;
}
```