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)

# https://qiita.com/norioc/items/5c270ad6ff5d22d1efd0#comment-da13f6b749dba662ec63
require "prime"

def divisors(n)
  n.prime_division.inject([1]) do |ary, (p, e)|
    (0..e).map{ |e1| p ** e1 }.product(ary).map{ |a, b| a * b }
  end.sort
end

N = gets.to_i
A = gets.split.map &:to_i

boxes = [0] * N
ans = [0] * N

(1..N).reverse_each{|i|
  # p [i-1, A[i-1],ans[i-1]]
  next if A[i-1] == boxes[i-1]
  ans[i-1] = 1
  divisors(i).each{|j|
    boxes[j-1] = boxes[j-1] + 1 & 1
  }
  # p [divisors(i), boxes]
}
p ans.size
puts ans * " "