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 OCaml (4.02.3) (judge verdict: Accepted)

let readInt () = Scanf.scanf " %d" (fun x -> x)

let solve n a =
  let b = Array.make n (-1) in
  let rec go i =
    if i < 1 then ()
    else
    let rec sumPrev s j =
      if j > n then s else sumPrev (s + b.(j-1)) (j + i)
    in
    let tot = sumPrev 0 (2*i) mod 2 in
    (*Printf.printf "%d %d %d\n" i tot a.(i-1);*)
    b.(i-1) <- if tot = a.(i-1) then 0 else 1;
    go (i-1)
  in go n;
  let rec go (k, acc) i =
    if i < 1 then (k, acc)
    else go (if b.(i-1) = 1 then (k+1, i :: acc) else (k, acc)) (i-1)
  in go (0, []) n

let main () =
  let n = readInt () in
  let a = Array.init n (fun _ -> readInt ()) in
  let (k, ans) = solve n a in
  Printf.printf "%d\n" k;
  ans |> List.iter (fun x -> Printf.printf "%d " x);
  print_endline ""

let () = main ()