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 Lua (5.3.2) (judge verdict: Accepted)

local N = io.read("n")
local a = 0
for i=1,N do
    if i < 10 then
        a = a + 1
    elseif i >= 100 then
        if i < 1000 then
            a = a + 1
        elseif i >= 10000 then
            if i < 100000 then
                a = a + 1
            end
        end
    end
end
print(a)