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: Runtime Error)

#include <stdio.h>
int main(void) {
	int n, c, a[100000], b[100000];
	int i, j;
	scanf("%d", &n);
	for (i = 0; i < n; i++) {
		scanf("%d", &a[i]);
	}
	
	for (i = n / 2; i < n; i++) {
		if (a[i] == 1) {
			b[i] = i + 1;
		}
		else if (a[i] == 0) {
			b[i] = 0;
		}
	}
	for (i = 1; i <= n / 2; i++) {
		c = 0;
		for (j = n / 2; j > 0; j = j - i) {
			if (b[i - 1] != 0) {
				c = c + 1;
			}
		}
		if (c % 2 == 0) {
			if (a[i - 1] == 1) {
				b[i - 1] = i;
			}
			else if (a[i - 1] == 0) {
				b[i - 1] = 0;
			}
		}
		else if (c % 2 == 1) {
			if (a[i - 1] == 0) {
				b[i - 1] = i;
			}
			else if (a[i - 1] == 1) {
				b[i - 1] = 0;
			}
		}
	}
	c = 0;
	for (i = 0; i < n; i++) {
		if (b[i] != 0) {
			c = c + 1;
		}
	}

	printf("%d\n", c);
	for (i = 0; i < n; i++) {
		if (b[i] != 0) {
			printf("%d\n", b[i]);
		}
	}
	return 0;
}