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 C (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>

#define N 200000

int main() {
	static int aa[N], bb[N];
	int n, i, j, x, y, m;

	scanf("%d", &n);
	for (i = 0; i < n; i++)
		scanf("%d", &aa[i]);
	for (i = n - 1; i > 0; i--)
		if (aa[i] == 1) {
			j = i + 1;
			for (x = 1; x * x <= j; x++)
				if (j % x == 0) {
					y = x - 1;
					aa[y] = (aa[y] + 1) % 2;
					if (x > 1 && x * x != j) {
						y = j / x - 1;
						aa[y] = (aa[y] + 1) % 2;
					}
				}
		}
	m = 0;
	for (i = 0; i < n; i++)
		if (aa[i] == 1)
			bb[m++] = i;
	printf("%d\n", m);
	for (i = 0; i < m; i++)
		printf("%d ", bb[i] + 1);
	if (m > 0)
		printf("\n");
	return 0;
}