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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int a[] = new int[n + 1];//添え字で１～ｎまでにしているので、１つ要素を多く持たせる
		int b[] = new int[n + 1];
		int cnt = 0;

		//pos以降はa[i]をb[i]にコピーすればよい（pos以降のiの倍数は１つしか存在しないため、a[i]の値と一致する）
		int pos = n / 2 + n % 2;
		for (int i = 1; i <= n; i++) {
			a[i] = sc.nextInt();

			if (i > pos) {
				b[i] = a[i];
				if (a[i] == 1)
					cnt++;
			}
		}
		for (int i = pos; i >= 1; i--) {
			int ballCnt = 0;
			for (int j = i; j <= n; j += i)
				ballCnt += b[j];
			if (ballCnt % 2 != a[i]) {
				b[i] = 1;
				cnt++;
			}
		}
				System.out.println(cnt);
		for (int i = 1; i <= n; i++) {
			if (b[i] == 1)
				System.out.println(i);
		}
	}
}