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;
#define all(a)a.begin(),a.end()
using ll=long long;
const int INF = 1<<30;
const ll INFll =1LL<<62;
const int mod= int(1e9)+7;
using P = pair<ll,ll>;
using ld=long double;

int main(){
    ll n;cin >>n;
    vector<ll>a(n);
    for (int i = 0; i < n; ++i) {
        cin >>a[i];
    }
    vector<ll>b;
    for (int i = n-1; i>=0;--i) {
        if(a[i]==1){
            b.push_back(i+1);
            a[0]=1-a[0];
            for (int j = 2; j*j<=(i+1); ++j) {
                if((i+1)%j==0){
                    a[j-1]=1-a[j-1];
                    if(j*j!=(i+1)) {
                        a[((i + 1) / j) - 1] = 1 - a[((i + 1) / j) - 1];
                    }
//                    cout <<j-1<<endl;
                }
            }
        }
    }
    ll m=b.size();
    cout <<m<<endl;
    for(auto e:b){
        cout <<e<<" ";
    }
    cout <<endl;
    return 0;
}