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

int main() {
int N;
cin >> N ;
int cnt=0;
for (int i = 1; i <=N; i++)
{
  
  string x = to_string(i);
  if(x.size()%2!=0) cnt++;
}
cout << cnt << endl;
  return 0;
}
### Prompt

Your task is to create a Cpp solution to 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++)
{
  
  string x = to_string(i);
  if(x.size()%2!=0) cnt++;
}
cout << cnt << endl;
  return 0;
}
```