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: Accepted)

/**
*    author:  souzai32
*    created: 14.08.2020 15:16:03
**/

#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
using namespace std;

int main() {

    int n;
    cin >> n;
    vector<int> a(n+1);
    for(int i=1; i<=n; i++) cin >> a.at(i);
    vector<int> ans;

    for(int i=n; i>0; i--){
        if(a.at(i)%2){
            ans.push_back(i);
            for(long long j=1; j*j<=i; j++){
                if(i%j==0){
                    a.at(j)++;
                    if(j*j!=i) a.at(i/j)++;
                }
            }
        }
        // rep(j,ans.size()-1) cout << ans.at(j) << ' ';
        // cout << endl;
    }
    cout << ans.size() << endl;
    reverse(ans.begin(),ans.end());
    rep(i,ans.size()-1) cout << ans.at(i) << ' ';
    if(ans.size()) cout << ans.at(ans.size()-1) << endl;

    return 0;
}