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 long long int ll;
#define rep(i,n) for(ll i=0;i<(n);i++)

int main(){
  ll n;cin>>n;
  vector<ll> v(n);
  vector<ll> w(n,0);
  rep(i,n) cin>>v[i];
  for(ll i=n-1;i>=0;i--){
      ll tmp=0;
      rep(j,n/(i+1)){
        if(j==0) tmp+=v[(i+1)*(j+1)-1];
        else tmp+=w[(i+1)*(j+1)-1];
      }
        if(tmp%2==0) w[i]=0;
        if(tmp%2==1) w[i]=1;
  }
  ll ans=0;
  rep(i,n) if(w[i]==1) ans++;
  cout<<ans<<endl;
  rep(i,n) if(w[i]==1) cout<<i+1<<' ';
  cout<<endl;
}