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 Kotlin (1.0.0) (judge verdict: Wrong Answer)

fun main(args : Array<String>) {
    val N = readLine()!!.toInt()
    val a = readLine()!!.split(" ").map { it.toInt() }.toIntArray()
    val b = IntArray(N)
    // O(N)
    (N - 1 downTo 0).forEach {
        val step = it + 1
        b[it] = a[it]
        // O(√N)
        val n = it
        for (i in 1..(Math.sqrt(n.toDouble())).toInt()) {
            if (n % i == 0) {
                a[i - 1] = (a[i - 1] + b[it]).mod(2)
                if (i != n / i) {
                    a[n/i - 1] = (a[n/i - 1] + b[it]).mod(2)
                }
            }
        }
    }

    val ans = b.mapIndexed { index, i -> if (i == 1) index + 1 else - 1 }.filter { it != -1 }
    println(ans.size)
    ans.forEach {
        print(it)
        print(' ')
    }
}