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>
int main(void){
  int n;
  scanf("%d", &n);
  int a[n];
  int b[n];
  int n_right = -1;

  for(int i=0;i<n;i++) {
  	scanf("%d", &a[i]);
  }

  int s;
  for(int i=n-1;i>=0;i--){
    b[i] = 0;
    s = 0;
    //s: kがiの倍数のb[k]の和
    //a_2 b_2,4,6,8
    //i:1 k:1,3,5,7 k+=(i+1)
    //a_1 b_1,2,3,4
    //i:0 k:0,1,2,3 k+=(i+1)

    for(int k=i; k<n; k+=i+1){
      s += b[k];
    }
    unsigned int tmp_s = s&1;
    unsigned int tmp_a = a[i]&1;
    if(tmp_s != tmp_a){
    	b[i] = 1;
    	if(n_right<0) n_right = i;
    } 
  }
  
  if(n_right<0){
    printf("0");  	
  }else if(n_right==0){
  	printf("1\n%d", b[0]);
  }else{
	printf("%d\n", n_right+1);
  	for(int i=0; i<n_right; i++){
  		printf("%d ", b[i]);
    }
  		printf("%d", b[n_right]);    
  }
  


    
  return 0;
}