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

int main(){
    int n,a=0,ans=0;
    cin>>n;

    for(int i=1; i<=n; i++){
        a=(int)log10((double)i)+1;
        if(a%2==1) ans++;
    }
    cout<<ans<<endl;
}

### 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 main(){
    int n,a=0,ans=0;
    cin>>n;

    for(int i=1; i<=n; i++){
        a=(int)log10((double)i)+1;
        if(a%2==1) ans++;
    }
    cout<<ans<<endl;
}

```