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: Accepted)

#include <stdio.h>

int main(){
	int n;
	scanf("%d",&n);
	int a[n];
	int b[n];
	for(int i=0; i<n; i++){
		scanf("%d",&a[i]);
		b[i]=0;
	}
	
	for(int i=n; i>=1; i--){
		int sum=0;
		/*if((sum%2)==a[i-1]){
			printf("%d\n",sum);
			for(int j=1; j<=n; j++){
				int flg=0;
				if(a[j-1]==1){
					if(flg==0){	
						printf("%d",j);
						flg ++;
					}else{
						printf(" %d",j);
					}
				}
			}
		}*/
		for(int j=1; j<=(n/i); j++){
			sum+=b[j*i-1];
		}
		b[i-1]=(sum+a[i-1])%2;
	}
	
	int sum_b=0;
	for(int i=0; i<n; i++){
		sum_b+=b[i];
	}
	
	printf("%d\n",sum_b);
	
	for(int j=1; j<=n; j++){
		int flg=0;
		if(b[j-1]==1){
			if(flg==0){	
				printf("%d\n",j);

		}
	}
}
}