Problem:
Score : 400 points

Problem Statement

There are N empty boxes arranged in a row from left to right.
The integer i is written on the i-th box from the left (1 \leq i \leq N).

For each of these boxes, Snuke can choose either to put a ball in it or to put nothing in it.

We say a set of choices to put a ball or not in the boxes is good when the following condition is satisfied:

For every integer i between 1 and N (inclusive), the total number of balls contained in the boxes with multiples of i written on them is congruent to a_i modulo 2.

Does there exist a good set of choices? If the answer is yes, find one good set of choices.

Constraints

All values in input are integers.

1 \leq N \leq 2 \times 10^5

a_i is 0 or 1.

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 ... a_N

Output

If a good set of choices does not exist, print -1.

If a good set of choices exists, print one such set of choices in the following format:

M
b_1 b_2 ... b_M

where M denotes the number of boxes that will contain a ball, and b_1,\ b_2,\ ...,\ b_M are the integers written on these boxes, in any order.

Sample Input 1

3
1 0 0

Sample Output 1

1
1

Consider putting a ball only in the box with 1 written on it.

There are three boxes with multiples of 1 written on them: the boxes with 1, 2, and 3. The total number of balls contained in these boxes is 1.

There is only one box with a multiple of 2 written on it: the box with 2. The total number of balls contained in these boxes is 0.

There is only one box with a multiple of 3 written on it: the box with 3. The total number of balls contained in these boxes is 0.

Thus, the condition is satisfied, so this set of choices is good.

Sample Input 2

5
0 0 0 0 0

Sample Output 2

0

Putting nothing in the boxes can be a good set of choices.

Solution in Ruby (2.3.3) (judge verdict: Wrong Answer)

require 'pp'

$n = gets.to_i + 1

nums = [0] + gets.split.map(&:to_i)

# p nums

# res = Array.new($n, 0)

def count_right_ones(index, nums)
    mul = 1
    i = index * mul
    res = 0
    # puts "   count: start(#{index})"

    while i < $n - 1
        # puts "   count: nums[#{i}] = #{nums[i]}"
        res += nums[i]
        mul += 1
        i = index * mul  - 1
    end
    return res
end

res = []

nums.each_with_index.reverse_each do |num, i|
    next if i == 0
    # puts "==="
    ones = count_right_ones(i, nums)
    ones_mod = ones % 2
    if ones_mod != num
        # puts ones
        # p [i, ones_mod, num]
        # p nums
        nums[i] = 1
        # p nums
        res << i
    end
    # puts "---"        
end

puts res.size
puts res.join(' ') if res.size > 0