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.to_i
if n <= 9
  puts n
  exit
elsif 10 <= n && n <= 99
  puts 9
  exit
elsif 100 <= n && n <= 999
  puts 9 + (n-100+1)
  exit
elsif 1000 <= n && n <= 9999
  puts 9 + 900
  exit
elsif 10000 <= n && n <= 99999
  puts 9 + 900 + (n-10000+1)
  exit
else
  puts 90909
  exit
end