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: Runtime Error)

n=gets.chomp.to_i
s = 1
max_value = {"s1": 9, "s3": 900}
kq = 0
while n/10 > 0
  n = n /10
  s+= 1
end
if(s%2==0)
  kq = maxvalue["s"+(s-1).to_s]
else
  kq = maxvalue["s"+(s-1).to_s] + n - 10**s
puts kq