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>

#pragma GCC optimize ("O3")
#pragma GCC target ("avx")

int b[200000];
int tmp,ans;

int main(void){
    int n;
    register int i,j;
    scanf("%d",&n);
    int a[n];
    for(i = 0;i < n;i++)scanf("%d",&a[i]);
    int tmp = 0;
    b[n-1]= a[n-1];
    for(i = n-2;i >= 0;i--){
        int add = 1;
        for(j = i;j < n;j += add)if(!((j+1) % (i+1))){
            tmp += b[j];
            add = i+1;
        }
        tmp %= 2;
        b[i] = tmp == a[i]?0:1;
    }
    for(i = 0;i < n;i++)ans += b[i];
    printf("%d\n",ans);
    for(i = 0;i < n;i++)if(b[i])printf("%d ",i+1);
    return 0;
}