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)

def IL(): return list(map(int,input().split()))
def SL(): return input().split()
def I(): return int(input())
def S(): return list(input())
import copy
def divisor(n): #約数
    f=[]
    d=[1]
    c=0
    r=int(n**0.5)
    for i in range(2,r+2):
        while n%i==0:
            c+=1
            n=n//i
        if c!=0:
            f.append([i,c])
            c=0
    if n!=1:
        f.append([n,1])
    for i in range(len(f)):
        t=[]
        for j in range(1,f[i][1]+1):
            t.append(f[i][0]**j)
        for j in range(len(d)):
            for k in range(len(t)):
                d.append(d[j]*t[k])
    return sorted(d)
n=I()
A=IL()
Ans=copy.deepcopy(A)
AA=[0]*n
AAA=[0]*n
last=-1
first=n
ans=0
c=True
while first>0:
    last=first
    first=first//2
    for i in range(first,last):
        if A[i]==1:
            if AA[i]==1:
                c=False
                break
            AA[i]=1
            L=divisor(i+1)
            for l in L:
                if A[l-1]==0:
                    A[l-1]=1
                else:
                    A[l-1]=0
                    
                if AAA[l-1]==0:
                    AAA[l-1]=1
                else:
                    AAA[l-1]=0   
                

if AAA!=Ans or not c:
    print(-1)
    print(A)
else:
    print(sum(AA))
    for i in range(n):
        if AA[i]==1:
            print(i+1,end=" ")
    print()