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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <math.h>

int divisor(int num,int d[])
{
	int i,j=0,c[(int)sqrt(num)+10],k,e=0;
	int castle=0;
	for(i=1;i<=(int)sqrt(num);i++)
	{
		if(num%i==0)
		{
			d[j] = i;j++;
			if(i != num/i){
				c[e] = num/i;
				e++;
			}
		}
	}
	k = e;
	for(i = j; i < j+e;i++)
	{
		k--;
		d[i] = c[k];
	}
	/*
	for(i = 0; i < j+e; i++)
	{
		printf("d[%d] = %d\n",i,d[i]);
	}
	*/
	return (e+j);
}

int main(void)
{
	int N;
	scanf("%d",&N);
	int i,a[N+1],b[N+1],ans[N+1],M=0,c[N+1],kate,j;
	for(i=1; i < N+1 ; i++)
	{
		scanf("%d",&a[i]);
		c[i] = 0;		b[i] = 0;		ans[i] = 0;
	}
	for(i=N; i > 0; i--)
	{
		kate = 0;
		if(a[i] == 0 && b[i] % 2 == 1)
		{
			kate = divisor(i,c);
			ans[i] = 1;
		}
		else if(a[i] == 1 && b[i] % 2 == 0)
		{
			kate = divisor(i,c);
			ans[i] = 1;
		}
		for(j = 0; j < kate; j++)
		{
			if(b[c[j]] == 1) b[c[j]]++;
			else if(b[c[j]] == 0) b[c[j]]++;
		}
	}
	for(i = 1; i < N+1; i++)
	{
		if(b[i] == 1) M++;
	}
	
	printf("%d\n",M);
	for(i = 1; i < N+1; i++)
	{
		if(ans[i] == 1) printf("%d ",i);
	}
	return 0;
}