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 (Clang 3.8.0) (judge verdict: Wrong Answer)

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

int main(){
    int N,A[200001],B[200001];
    int i,M;
    M = 0;
    scanf("%d",&N);
    for(i = 0 ; i < N ; i++){
        scanf("%d",&A[i]);
    }
    int counter = 1;
    int ball_num = 0;
    for(i = N - 1 ; i >= 0 ; i--){
        if(A[i] == 1){
            if(M < (i + 1)){
                M = i + 1;
            }
            while(1){
                if(counter * (i + 1) > N){
                    break;
                }else{
                    ball_num += B[i];
                }
                counter++;
            }
            counter = 0;
            if(ball_num % 2 == 0){
                B[i] = 1;
            }
        }
    }
    printf("%d\n",M);
    for(i = 0;i<M;i++)
        printf("%d ",B[i]);
    return 0;
}