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)

def test(n)
  return 90909 if n == 100000
  return (10000..n).size + 909 if n.between?(10000,99999)
  return 909 if n.between?(1000,9999)
  return (100..n).size + 9 if n.between?(10,999) 
  return (1..n).size if n.between?(1,9) 
end
puts test(gets.to_i)