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: Runtime Error)

import scala.io.StdIn

object Main {

  def divisors(n: Int): Vector[Int] = {
    def rec(i: Int, acc: Vector[Int]): Vector[Int] = {
      if (i * i > n) acc
      else if (i * i == n) acc :+ i
      else if (n % i == 0) rec(i + 1, i +: acc :+ (n / i))
      else rec(i + 1, acc)
    }
    rec(1, Vector.empty[Int])
  }

  def switch(as: Vector[Int], n: Int): Vector[Int] = {
    divisors(n).foldLeft(as)((acc, d) => as.updated(d, 1 - as(d)))
  }

  def solve(as: Vector[Int], n: Int): Vector[Int] = {
     if (n == 0) Vector(0)
     else {
       if (as(n) == 0) {
         solve(as.dropRight(1), n - 1) :+ 0
       } else {
         solve(switch(as, n), n - 1) :+ 1
       }
     }
  }

  def main(args: Array[String]): Unit = {
     val n = StdIn.readInt()
     val as = 0 +: StdIn.readLine().split(' ').map(_.toInt).toVector
     val bs = solve(as, n)
  }

}