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: Runtime Error)

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
#define int long long
#define rep(i,n) for(int i = 0; i < (n); i++)
#define divup(a,b) ( ( (a) + ( (b) - 1u) ) / (b) ) // a/b の余り切り上げ
// #include <boost/multiprecision/cpp_int.hpp>
// using namespace boost::multiprecision;
ll mod = 1000000007;

signed main(){
    ll n;
    ll a[200009];
    ll b[200009];
    ll k[200009];
    vector<ll> ss;
    cin>>n;
    rep(i,n) cin>>a[i];
    rep(i,n) b[i]=0,k[i]=0;
    ll sk=1;
    bool f=false;
    rep(i,n){
        ll ni=n-1-i;
        if(ni!=0){
            if(ni<(n/pow(2,sk))){
                sk++;
            }
            if(ni>=n/pow(2,sk)){
                rep(j,pow(2,sk-1)){
                    if(k[(ni+1)*(j+1)-1]==1) b[ni]++;
                }
                if(a[ni]!=(b[ni]%2)){
                    k[ni]=1;
                }
            }
        }else{
            rep(j,n){
                if(k[j]==1) b[ni]++;
            }
            if(a[ni]!=(b[ni]%2)){
                k[ni]=1;
            }
        }
    }
    ll ks=0;
    rep(i,n) ks += k[i]; 
    cout<<ks<<endl;
    vector<ll> kk;
    rep(i,n){
        if(k[i]==1) kk.push_back(i+1);
    }
    ll sss=kk.size();
    rep(i,sss){
        if(i!=sss-1) cout<<kk[i]<<" ";
        else cout<<kk[i]<<endl;
    }
}