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)

import java.util.*
fun main(args: Array<String>){
	val cin = Scanner(System.`in`)
    val n = cin.nextInt()
    val a = Array<Int>(n, { cin.nextInt() })
    val ans = Array<Int>(n, { 0 })
    var sum = 0

    val prime: MutableList<Int> = mutableListOf()

    for (i in 2..n) {
        var ok = true
        for (p in prime) {
            if (i % p == 0) {
                ok = false
                break
            }
        }
        if (ok) {
            prime.add(i)
        }
    }

    for (i in 0..n-1) {
        val j = n-i //番号
        if (j*2 > n) {
            ans[j-1] = a[j-1]
            if (a[j-1] == 1) {
                sum += 1
            }
        } else if (j == 1) {
            if (a[j-1] != a[2*j-1]) {
                ans[j-1] = 1
                sum += 1
            }
        } else {
            var s = 0
            for (t in prime) {
                if (t*j > n) {
                    break
                }
                s += ans[t*j-1]
            }
            if (s % 2 != a[j-1]) {
                ans[j-1] = 1
                sum += 1
            }
        }
    }

    println(sum)
    for (i in 0..n-1) {
        if (ans[i] == 1) {
            println(i+1)
        }
    }
}