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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
int main(void) {
	int n;
	static int a[200000], b[200000]={0};
	int half, co,sum;
	int i, j;

	scanf("%d", &n);
	for (i = 0; i < n; i++) {
		scanf("%d", &a[i]);
	}
	/*半分までの確定している部分をb[i]に入れる*/
	if (n % 2 == 0) {
		half = n / 2;
	}
	for (i = half; i < n; i++) {
		b[i] = a[i];
	}
	/*半分よりまえの部分をb[i]に入れる*/
	for (i = half; i >= 0; i--) {
		sum = 0;//カウントリセット
		for (j = i; j < n; j += i+1) {
			if (b[j] == 1) {
				sum+=b[j];//ボールの数の合計
			}
		}
		if(sum%2!=a[i]){
			b[i]=1;
		}
	}
	co = 0;
    for(i=0;i<n;i++){
        if(b[i]==1){
			co++;//1の数をカウント
		}
    }
	printf("%d\n",co);
	int cnt=0;
	for(i=0;i<n;i++){
		if(b[i]==1){
			printf("%d ",i+1);
		}
	}
	return 0;
}