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

Create a solution in CPP for 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;
    cin >> n;
    int cnt = 0;
    for(int i=1;i<=n;i++){
        if((int)log10(i) % 2 == 0) cnt++;
    }
    cout << cnt << endl;
}
```