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

arr = N.to_s.split('')

count = 0
l = arr.length
if l == 1
  print 0
elsif l > 1
  count += 9
else
  return print N
end

if l == 3
  count += N - 99
end

if l > 3
  count += 900
end

if l == 5
  count += N - 9999
end

if l > 5
  count += 90000
end

print count