#include<bits/stdc++.h>
using namespace std;
int main()
{

long long a,b,c=0,d,e,g,h,i,j,k;
string arr;
cin>>a;
for(i=1;i<=a;i++)
{
arr=to_string(i);
if(arr.length()%2==1)
c++;

}
cout<<c;




}

### 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()
{

long long a,b,c=0,d,e,g,h,i,j,k;
string arr;
cin>>a;
for(i=1;i<=a;i++)
{
arr=to_string(i);
if(arr.length()%2==1)
c++;

}
cout<<c;




}

```