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>
#include <vector>
#include <numeric>
#define PI 3.14159265358979323846
#define MAXINF (1e18L)
#define INF (1e9L)
#define EPS (1e-9)
#define MOD ((ll)(1e9+7))
#define REP(i, n) for(int i=0;i<int(n);++i)
#define Rep(i,sta,n) for(int i=sta;i<n;i++)
#define RREP(i, n) for(int i=int(n)-1;i>=0;--i)
#define ALL(v) v.begin(),v.end()
#define FIND(v,x) (binary_search(ALL(v),(x)))
#define SORT(v) sort(ALL(v))
#define RSORT(v) sort(ALL(v));reverse(ALL(v))
#define DEBUG(x) cerr<<#x<<": "<<x<<endl;
#define DEBUG_VEC(v) cerr<<#v<<":";for(int i=0;i<v.size();i++) cerr<<" "<<v[i]; cerr<<endl
#define Yes(n) cout<<((n)?"Yes":"No")<<endl
#define YES(n) cout<<((n)?"YES":"NO")<<endl
#define pb push_back
#define fi first
#define se second
using namespace std;
template<class A>void pr(A a){cout << (a) << endl;}
template<class A,class B>void pr(A a,B b){cout << a << " "  ;pr(b);}
template<class A,class B,class C>void pr(A a,B b,C c){cout << a << " " ;pr(b,c);}
template<class A,class B,class C,class D>void pr(A a,B b,C c,D d){cout << a << " " ;pr(b,c,d);}
template<class T> inline bool chmin(T& a, T b){return a>b ? a=b, true : false;}
template<class T> inline bool chmax(T& a, T b){return a<b ? a=b, true : false;}
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll,ll> pll;

int main(void)
{
    int N;
    cin >> N;
    vector<int> a(N+1);
    for (int i = 1; i <= N; i++) {
        cin >> a[i];
    }
    vector<int> flag(N,0);
    set<int> ans;
    int t = 0;
    for (int i = N; i > 0; i--){
        if(flag[i-1]) {
            int tmp = i*2;
            int count = 0;
            while(tmp <= N){
                if(ans.find(i) != ans.end())count++;
                tmp+=i;
            }
            if((count&1) != (t&1)){
                pr(-1);
                return 0;
            }
        }
        if((a[i]==0&&(t&1)==1) || (a[i]==1&&(t&1)==0)){
            ans.insert(i);
            t++;
        }
        int tmp = i;
        while((tmp&1)==0){
            flag[tmp-1] = 1;
            tmp /= 2;
        }
    }
    if(ans.empty()){
        pr(0);
        return 0;
    }
    pr(ans.size());
    for(auto&& i : ans) pr(i);

    // vector<int> ans;
    // for (int i = N; i > 0; i--){
    //     if(i*2>N){
    //         if(a[i]) ans.pb(i);
    //     }else{
    //         if(a[i*2]==1 && a[i]==0)ans.pb(i);
    //         else if(a[i*2]==0 && a[i]==1)ans.pb(i);
    //     }
    // }
    // if(ans.empty()){
    //     pr(0);
    //     return 0;
    // }
    // SORT(ans);
    // int n = ans.size();
    // pr(n);
    // REP(i, n){
    //     cout << ans[i] << " ";
    // }
    // cout << endl;
}