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 9.2.1) (judge verdict: Wrong Answer)

/*
"Won't stop until I'm phenomenal."
- Phenomenal, Eminem
 */
#include<bits/stdc++.h>
using namespace std;
typedef long long int ll;
#define ff first
#define Shazam ios_base::sync_with_stdio(false);   cin.tie(NULL); cout.tie(NULL);
#define ss second
#define all(c) c.begin(),c.end()
#define endl "\n"
#define test() int t; cin>>t; while(t--)
#define fl(i,a,b) for(int i = a ; i <b ;i++)
#define get(a) fl(i,0,a.size())  cin>>a[i];
#define pra(a) fl(i,0,a.size()) cout<<a[i]<<" ";	cout<<endl;
#define pr(a,n) fl(i,0,n) cout<<a[i]<<" ";	cout<<endl;
const ll INF = 2e18;
const int inf = 2e9;
const int mod1 = 1e9 + 7;



int main(){
    Shazam;
    int n; cin>>n;
    vector<int> a(n+1);
    for(int i = 1; i <= n; i++) cin>>a[i];
    int res = 0;
    vector<int> ans(n+1);
    for(int i = n; i>0; i--){
        int curr = 0;
        for(int j = 2*i; j <=n; j+=i){
            curr+=ans[j];
        }
        if ((curr & 1) != (a[i] & 1)){
            ans[i] = 1;
            res++;
        }
    }
    cout<<res<<endl;
    for(int i : ans){
        if(i)
            cout<<res<<endl;
    }
  
    return 0;
}