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: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
using ll = long long;

template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }

#define FOR(i,a,b) for(ll i=(a);i<(b);++i)
#define ALL(v) (v).begin(), (v).end()
#define p(s) cout<<(s)<<endl
#define p2(s, t) cout << (s) << " " << (t) << endl
#define br() p("")
#define pn(s) cout << (#s) << " " << (s) << endl
#define p_yes() p("YES")
#define p_no() p("NO")

const ll mod = 1e9 + 7;
const ll inf = 1e18;

void vprint(vector<ll> A){
    ll L = A.size();
    FOR(i, 0, L){
        if(i) cout << ' ';
        cout << A[i];
    }
    cout << endl;
}

vector<ll> calc_devisors(ll a){
    vector<ll> ret;
    for(int i=1; i*i<=a; i++){
        if(a%i!=0) continue;
        ret.push_back(i);
        ll another = a/i;
        if(i!=another) ret.push_back(another);
    }
    return ret;
}

const int N_MAX = 200010;
ll C[N_MAX];

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);

    // input
    ll N;
    cin >> N;

    vector<ll> A(N);
    FOR(i, 0, N){
        cin >> A.at(i);
    }

    vector<ll> Ans(N);

    for(int i=N-1; i>=0; i--){
        // 今まで何個入っているか
        ll c = C[i];

        ll a = A[i];
        if((a==1 && c==0) || (a==0 && c==1)){
            Ans[i]=1;
            // いれる
            auto V = calc_devisors(a);
            for(ll v : V){
                C[v]++;
                C[v] %= 2;
            }

        }else{
            // いれない
        }

    }

    ll count=0;
    for(ll a : Ans){
        if(a==1) count++;
    }
    p(count);
    FOR(i, 0, N){
        if(Ans[i]==1){
            p(i+1);
        }
    }
    
    return 0;
}