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 < 10 
    puts n # 1..n
elsif n < 100
    puts 9 # 1..9
elsif n < 1000
    puts n - 90 # 1..9, 100..n
elsif n < 10000
    puts 909 # 1..9, 100..999
elsif n < 100000
    puts n - 9090 # 1..9, 100..999, 10000..n
elsif n == 100000
    puts 90909 # 1..9, 100..999, 10000..99999
end