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)

from collections import defaultdict
N = int(input())
arr = list(map(int, input().split()))
puts = []
dd = defaultdict(int)

def make_divisors(n):
    divisors = []
    # Nの最大の約数は√N
    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)
    # divisors.sort()
    return divisors

for i in range(1,N+1)[::-1]:
    if N//2 < i and arr[i-1] == 1:
        puts.append(i)
        ds = make_divisors(i)
        for j in ds:
            dd[j] += 1
    else:
        # c = len([p for p in puts if p % i == 0])
        c = dd[i]
        if (c % 2 == 0 and arr[i-1] == 1) or (c % 2 == 1 and arr[i-1] == 0):
            puts.append(i)
            ds = make_divisors(i)
            for j in ds:
                dd[j] += 1

print(len(puts))
if len(puts):
    print(" ".join(map(str, sorted(puts, reverse=True))))