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 PyPy3 (2.4.0) (judge verdict: Accepted)

N = int(input())
a = list(map(int, input().split()))
 
def make_divisors(n):
    divisors = []
    for i in range(1, int(n**0.5)+1):
        if n % i == 0:
            divisors.append(i)
            if i != n // i:
                divisors.append(n//i)
    return divisors
 
 
tmp = [0]*(N+1)
b = [0]*N
ball_num = 0
ball_box = []
for i in reversed(range(N)):
    if a[i] == 0:
        if (tmp[i+1]%2) == 0:
            continue
        else:
            ball_box.append(i+1)
            ball_num += 1
            val = make_divisors(i+1)
            for elms in val:
                tmp[elms] += 1
    elif a[i] == 1:
        if (tmp[i+1]%2) == 0:
            ball_box.append(i+1)
            ball_num += 1
            val = make_divisors(i+1)
            for elms in val:
                tmp[elms] += 1
        else:
            continue
 
print(ball_num)
if ball_num != 0:
    print(*ball_box)