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: Runtime Error)

n = gets.chomp.split('').map(&:to_i)
number = 0

for i in 1..n.length
    if n.length == i
        if i % 2 == 1
            for num in 0..i
                number += n[num] * (10 ** (i - num))
            end
            i = i - 2 > 0 ? i : 0
            puts number - 9 * (10 ** i)
        else
            puts 9 * (10 ** (i - 2))
        end
        break
    end
end