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 C++14 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include<stdio.h>
int n,sum,m=0,t,h=0,c=0;
	int a[200008],b[200008];
int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	{
	scanf("%d",&a[i]);
//	if(a[i]==0)
//	h++;
   }
//if(h==n)
//printf("%d\n",0);
//else
//{
	for(int i=1;i<=n;i++)
	{
		sum=0;
		for(int j=1;j*i<=n;j++)
		sum+=a[j*i];
		while(sum!=0)
		{
			t=sum%2;
			sum=sum/2;
		}
		if(t==a[i])
		{
			m=1;
				for(int j=1;j*i<=n;j++)
				if(a[j*i]==1)
				{
					b[c]=i*j;
					c++;
				}
					break;
			}
		}
	if(m==0)
	printf("%d\n",-1);
	else
	{
	printf("%d\n",c);
	for(int i=0;i<c;i++)
	printf("%d ",b[i]);
	printf("\n");
}
return 0;
}