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 PHP7 (7.0.15) (judge verdict: Accepted)

<?php
fscanf(STDIN, "%d", $N);
$a = array_map('intval', explode(' ', trim(fgets(STDIN))));
$b = array_pad(array(), $N, null);

for ($i = $N; $i > 0; $i--) {
	if ($N / 2 > $i) {
		$tmp = $a[$i * 2 - 1];
		$st = $i * 3;
		$dj = $i * 2;
	} else {
		$tmp = 0;
		$st = $i * 2;
		$dj = $i;
	}
    for ($j = $st; $j <= $N; $j += $dj) {
        $tmp += $b[$j - 1];
    }
    $b[$i - 1] = ($tmp - $a[$i - 1] + 10) % 2;
}
$ans = [];
for ($i = 0; $i < $N; $i++) {
	if ($b[$i] === 1 ) $ans[] = $i + 1; 
}
printf("%d\n", count($ans));
echo implode(' ', $ans);