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.to_i

keta = n.to_s.length
case keta
when 1 then
    puts n
when 2 then
    puts 9
when 3 then
    puts n - 99 + 9 
when 4 then
    puts 9 + 900
when 5 then
    puts n - 999 + 9 + 900
when 6 then
    puts 9 + 900 + 90000
end