#include<bits/stdc++.h>
using namespace std;
int main()
{
  int n,no;
  cin>>n;
  int d=0;
  for(int i=1;i<=n;i++)
  {
   no=floor(log10(i)+1);\
     if(no%2!=0)
     {
       d++;
     }

  }
  cout<<d;
}

### 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,no;
  cin>>n;
  int d=0;
  for(int i=1;i<=n;i++)
  {
   no=floor(log10(i)+1);\
     if(no%2!=0)
     {
       d++;
     }

  }
  cout<<d;
}

```