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>
#include <string.h>

int main (){
  /*  int a, b, c;
  char x,y,z;
  //  scanf("%c%c%c", &x, &y, &z);
  int N, a;
  scanf("%d %d",&N, &a);
  scanf("%d",&a);
  int d=(int)3.8;
  */
  int N;
  scanf("%d",&N);
  int array[N+1];
  int out[N+1];
  int enable_sum=0;
  for (int i=1; i<=N ; i++){
    scanf("%d",&array[i]);
  }
  for (int i=1; i<=N; i++){
    //printf("array[%d]=%d\n",i,array[i]);
  }


  for (int i=N; i>=1; i--){
    //    printf("============--\n");
    //    printf("i=%d\n",i);
    int sum_other = 0;
    int j = 2*i;
    while(j<=N){
      //      printf("add %d\n",j);
      sum_other += out[j];
      j+=i;
    }
    //    printf("sum_other=%d\n",sum_other);
    out[i] = (array[i]+sum_other)%2;
    if (out[i]==1){
      enable_sum++;
    }
    //    printf("out[%d]=%d\n",i,out[i]);
  }

  printf("%d\n",enable_sum);
  int counter=0;
  for (int i=1; i<=N; i++){
    if (counter==0 & out[i]==1){
      printf("%d",i);
      counter=1;
    }
    else if (out[i]==1){
      printf(" %d",i);
    }
  }

  if (enable_sum!=0){
    printf("\n");
  }

  return 0;
}