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 (gfortran v4.8.4) (judge verdict: Accepted)

program abc136b
   implicit none
   integer :: n
   integer :: digit, counter

   read(*,*) n

   digit = int(log10(real(n))) + 1


   if ( digit == 6 ) then
      write(*,*) 90909
      stop
   endif

   counter = 0
   if ( digit == 5 ) then
      counter =  (n - 9999)
      counter = counter + 909
   else if (digit == 4) then
      counter = 909
   else if ( digit == 3) then
      counter =  (n - 99) + 9
   else if ( digit == 2) then
      counter = 9
   else if (digit == 1 ) then
      counter = n
   end if

   write(*,*) counter

end program abc136b