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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
typedef vector<int> vi;
typedef pair<int,int> pii;
typedef long long ll;
const long long MOD=1e09+7;



#define rep(i,n) for(int i=0;i<n;i++)
#define rep2(i,m,n) for(int i=m;i<n;i++)
#define ALL(v) v.begin(), v.end()





int main(){
  

  int n;
  cin>>n;
  vi data(n+1);
  for (int i=1;i<=n;i++){
    int tmp;
    cin>>tmp;
    data.at(i)=tmp;
  }
  
  vi answer(n+1,0);
  
  for (int i=n;i>=1;i--){
    int sum=0;
    for (int j=2*i;j<=n;j+=i){
      sum+=answer.at(j);
      sum%=2;
    }
    answer.at(i)=abs(data.at(i)-sum)%2;
  }
  
  int count=0;
  for (int i=1;i<=n;i++){
    count+=answer.at(i);
  }
  
  cout<<count<<endl;
  for (int i=1;i<=n;i++){
    if (answer.at(i)==1){
      cout<<i<<" ";
    }
  }
  cout<<endl;
  
  


}