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: Accepted)

n = gets.chomp.to_i

one = 9-1+1
thr = 999-100+1
fiv = 99999 - 10000 + 1
result = 0
case n.to_s.length
when 6 then
  result = one + thr + fiv
when 5
  result = one + thr + fiv - (99999 - n)
when 4 then
  result = one + thr
when 3
  result = one + thr - (999 - n)
when 2 then
  result = one
when 1 then
  result = one - (9 - n)
end
puts result