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 Scala (2.11.7) (judge verdict: Wrong Answer)

object Main extends App {
  val n = io.StdIn.readLine.toInt
  val as = io.StdIn.readLine.split(" ").map(_.toInt)
  var bs = new Array[Int](n)

  val bool = (n to 1 by -1).foldLeft(true)((b, i) => b match {
    case false => false
    case true => {
      val bb = (i * 2 to n by i).map(x => bs(x - 1)).sum & 1
      if (bb == 0) {
        bs(i - 1) = as(i - 1)
        true
      } else if (bb == 1 && as(i - 1) == 1) {
        bs(i - 1) = 0
        true
      } else {
        false
      }
    }
  })

  if (bool) {
    val bs2 = bs.zipWithIndex.filter(t => (t._1 & 1) == 1)
    println(bs2.length)
    println(bs2.map(t => t._2 + 1).mkString(" "))
  } else
    println(-1)
}