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 {
  var a = readInt
  var b = readLine.split(" ").map(_.toInt)
  var c,ans,s,tmp = 0
  var box = new Array[Int](a)
  
  for(i <- a-1 to 0 by -1) {
    c = i + 1
    for(j <- c - 1 until a by c) {
      tmp = tmp + box(j)
    }
    if(tmp %2!=b(i)) {
      ans = ans + 1
      box(i) = 1
    }
  }
  println(ans)
  for(i <- 0 until a) {
    if(box(i) == 1 && s != 0) {
      tmp = i + 1
      print(" " + tmp)
    }
    if(box(i) == 1 && s == 0) {
      tmp = i + 1
      print(tmp)
      s = s + 1
    }
  }
}