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
t = n
kq = 0
stt = t.to_s.length
if stt == 1
  kq = n
elsif(stt%2==0)
  for num in 1..(stt-1) do
    if num % 2 == 1
      kq += 10**(num-1)*9
    end
  end  
elsif(stt%2==1)
  for num in 1..(stt-2) do
    if num % 2 == 1
      kq += 10**(num-1)*9
    end
  end
  kq = n + 1 - 10**(stt-1)
end
puts kq