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: Time Limit Exceeded)

fun main(args: Array<String>) {
    val N = nextInt()
    val A = listOfInt().toMutableList()
    A.add(0, 0)
    val B = IntArray(N + 1) { if (it == 0) 0 else N / it}
    val C = IntArray(N + 1) { 0 }
    fun even(n: Int): Boolean = (n % 2 == 0)
    for (i in (1..N).reversed()) {
        var sum = 0
        for (n in (1..N).reversed()) {
            if (n % i == 0) sum += C[n]
        }
        if (A[i] == 0 && !even(sum)) C[i] = 1
        if (A[i] == 1 && even(sum)) C[i] = 1
    }
    for (i in 1..N) {
        var sum = 0
        for (n in 1..N) {
            if (n % i == 0) sum += C[n]
        }
        if (A[i] == 0 && even(sum)) continue
        if (A[i] == 1 && !even(sum)) continue
        println(-1)
    }
    var ans = C.filter { it  == 1}
    println(ans.count())
    println(ans.joinToString(separator = " "))
}
fun next() = readLine()!!
fun nextInt() = Integer.parseInt(next())
fun listOfString() = next().split(" ")
fun listOfInt(delta: Int = 0) = listOfString().map { Integer.parseInt(it) + delta }