#include <bits/stdc++.h>
using namespace std;
int main(){
int N,c=0;
cin >> N;
for(int i=1;i<=N;i++){
if((i<10)||(100<=i&&i<1000)||(10000<=i&&i<100000)){c++;}
}
cout << c;
}
### 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,c=0;
cin >> N;
for(int i=1;i<=N;i++){
if((i<10)||(100<=i&&i<1000)||(10000<=i&&i<100000)){c++;}
}
cout << c;
}
```