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>
int n,a[200001],yak[200001][100],yac[200001]={0};
int hab[200001]={0},b[200000]={0};

int main(){
	int i,m=0;
	scanf("%d",&n);
	for(i=1;i<=n;i++)scanf("%d",&a[i]);
	for(i=1;i<=n;i++){
		int j,f=0;
		yak[i][yac[i]++]=1;
		if(i==1)continue;
		for(j=2;j*j<=i;j++){
			if(i%j==0){
				yak[i][yac[i]++]=j;
			}
		}
		for(j=0;j<yac[i];j++){
			if(yak[i][j]*yak[i][j]!=i)yak[i][yac[i]+j]=i/yak[i][j];
			else f=1;
		}
		yac[i]*=2;
		if(f==1)yac[i]--;
	}
	for(i=n;i>0;i--){
		if(hab[i]%2!=a[i]%2){
			b[m++]=i;
			int j;
			for(j=0;j<yac[i];j++)hab[yak[i][j]]++;
		}
	}
	if(hab[1]%2!=a[1]%2)puts("-1");
	else{
		printf("%d\n",m);
		for(i=0;i<m;i++)printf("%d ",b[i]);
		puts("");
	}
	return 0;
}