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)

target=gets
number=target.to_i
keta = target.to_s.length.to_i
total=0
1.step(keta-1,2) do |num|
  total=total+(9*10**(num-1))
end
if keta%2 == 1
  total=total+(number-(10**(keta-1)))+1
end
puts total.to_i