Problem:
Score : 200 points

Problem Statement

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.

Sample Input 1

11

Sample Output 1

9

Among the positive integers less than or equal to 11, nine integers have an odd number of digits: 1, 2, \ldots, 9.

Sample Input 2

136

Sample Output 2

46

In addition to 1, 2, \ldots, 9, another 37 integers also have an odd number of digits: 100, 101, \ldots, 136.

Sample Input 3

100000

Sample Output 3

90909

Solution in Fortran (GNU Fortran 9.2.1) (judge verdict: Accepted)

program ABC136B
    implicit none
    integer::N,i,digits,j,ans=0
    read*,N
    do i=1,N
        j=i
        digits=1
        do
            if(j/10==0)exit
            j=j/10
            digits=digits+1
        end do
        if(mod(digits,2)==1)ans=ans+1
    end do
    print'(i0)',ans
end program ABC136B