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>
#include<stdlib.h>

#define NEW(p,n){p=malloc((n)*sizeof(p[0]));if(p==NULL){printf("not enough memory\n");exit(1);};}
//pの型の変数n個の要素分のメモリを確保し、そのアドレスをpに代入するマクロ

int main(void){
    int N;
    scanf("%d",&N);

    //A[1]~A[N]に格納する
    int* A;
    NEW(A,N+1);
    for(int i=1;i<=N;i++){
        scanf("%d",&A[i]);
    }

    //ボールを入れる箱の個数
    int M=0;

    //ボールを入れる方法を格納する
    //B[i]は番号iの箱
    int* B;
    NEW(B,N+1);
    for(int i=N;i>0;i--){
        //iの倍数が書かれた箱のボースの総数をsumに数える
        int index=i;
        int sum=0;
        while(index<=N){
            sum+=B[index];
            index+=i;
        }

        if((sum%2)==A[i]) B[i]=0;
        else{
            B[i]=1;
            M+=1;
        }
    }

    printf("%d\n",M);
    if(M!=0){
        for(int i=1;i<=N;i++){
            if(B[i]==1) printf("%d ",i);
        }
        printf("\n");
    }

    return 0;
}

//400点
//昇順で考えるより降順で考えたほうがいいこともあるよね