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>
#define FOR(n) for(int i=0;i<n;i++)
#define PRN(n) printf("%d\n",n)
#define PRS(s) printf("%s\n",s)
int a[200000];
int b[200000];
int main() {
  int n,cnt=0;
  scanf("%d",&n);
  FOR(n)scanf("%d",&a[i]);
  for(int i=n;i>0;i--){
    if(a[i-1]){
      for(int j=1;j<=i;j++){
        if(!(j%i)){
          a[i-1]++;
          a[i-1]%=2;
          b[i-1]=1;
          cnt++;
        }
      }
    }
  }
  PRN(cnt);
  FOR(n)if(b[i])printf("%d ",i+1);
  return 0;
}