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 (1.3.71) (judge verdict: Wrong Answer)

fun main() {
    abc134d()
}

fun abc134d() {
    val n = readLine()!!.toInt()
    // 各桁について、1ならiの倍数が奇数個になるようにするという意味
    val a = BooleanArray(n + 1) { false }
    for ((i, s) in readLine()!!.split(" ").withIndex()) {
        a[i + 1] = s == "1"
    }
    val ans = mutableSetOf<Int>()
    var i = n
    var depth = 0
    while (i > 1) {
        var j = i
        while (j * 2 > i) {
            if (depth == 0) {
                if (a[j]) ans.add(j)
            } else {
                var count = 0
                for (k in 1..depth) {
                    if (a[j * k]) count++
                }
                val isOdd = count % 2 == 1
                if (isOdd xor a[j]) ans.add(j)
            }
            j--
        }
        i = j
        depth++
    }
    if (a[1]) {
        if (ans.size % 2 == 0) ans.add(1)
    } else {
        if (ans.size % 2 == 1) ans.add(1)
    }
    println(ans.size)
    println(ans.joinToString(" "))
}