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 inputs = readLine()!!.split(" ").map { it.toInt() }

    // 何の倍数なのか判定を残さなくてはいけないのは大変では．．．
    // 効率的に残したい いや 毎回すべて判定でも行ける！

    var ans = IntArray(n) { 0 }
    var num = 0
    for (i in n-1 downTo 0) {
        var sum = 0

        for (j in i until n) {

            if (j % (i+1) == 0) {
                sum += ans[j]
            }

        }

        if(sum % 2 != inputs[i]) {
            ans[i] += 1
            num += 1
        }
    }

    println(num)

    if(num != 0) {
        val ansIndex = mutableListOf<Int>()
        for (i in 0 until n){
            if(ans[i] == 1) {
                ansIndex.add(i+1)
            }
        }

        val out = ansIndex.joinToString(" ")
        println(out)
    }
}