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.to_i

if  N/10 < 0
	puts N
elsif  N/10 < 10
	puts 9
elsif  N/100 <  10
	puts 9 + N- 99
elsif  N/1000  <  10
	puts  9 + 900
elsif  N/10000  <  10
	puts  909 + N - 9999
else
	puts  90909
end