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 Ruby (2.3.3) (judge verdict: Wrong Answer)

n = gets.chomp.to_i
digits = n.to_s.chars.size
if digits == 1
  puts n
elsif digits.even?
  ans = (1..(digits-1)).step(2).inject(0) do |sum, i|
    sum += 9 * (10 ** ( i - 1 ))
  end
  puts ans
else
  ans = (1..(digits-2)).step(2).inject(0) do |sum, i|
    sum += 9 * (10 ** ( i - 1 ))
  end + n - n.to_s.chars.fill(0,1).join().to_i + 1
  puts ans
end