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 (7.3.0) (judge verdict: Accepted)

from collections import Counter
N=int(input())
A=[0]+list(map(int,input().split()))

def fast_factorization_init(N):
    res=list(range(N))
    for i in range(2,N):
        if i*i>N:
            break
        for j in range(i*i,N,i):
            if res[j]==j:
                res[j]=i
        if res[i]<i:
            continue
    return res            
    
min_factors=fast_factorization_init(2*10**5+10)
def fast_factorization(n):
    res=[]
    while n>1:
        res.append(min_factors[n])
        n//=min_factors[n]
    return res

def factors(n):
    res=[1]
    for i,j in Counter(fast_factorization(n)).items():
        res+=[r*i**(b+1) for r in res for b in range(j)]
    return res

res=[0]*(N+1)
num=0
for n in reversed(range(N)):
    res[n+1]=A[n+1]
    if A[n+1]:
        num+=1
        for m in factors(n+1):
            A[m]=(A[m]+1)%2
print(num)
print(*[i+1 for i in range(N) if res[i+1]>0],sep='\n')