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 REP(i,m) for(ll i=0; i<m; i++)
#define FOR(i,n,m) for(ll i=n; i<m; i++)
#define INF 1000000007
#define INFL (1LL << 60)
#define MOD 1000000007
#define ALL(v) v.begin(),v.end()
#define SORT(v) sort(ALL(v))
#define pb push_back
#define ll long long int
#define P pair<ll, ll>

int main(){
    int n;
    cin >> n;
    int a[n];
    REP(i, n) cin >> a[i];
    int ans[n];
    REP(i, n){
        int x = n-i;
        int count = 0, cp = 2;
        while(cp*x-1 < n){
            count += ans[cp*x-1];
            cp++;
        }
        if(count%2 == a[x-1]) ans[x-1] = 0;
        else ans[x-1] = 1;
    }
    vector<int> b;
    REP(i, n) if(ans[i] == 1) b.pb(i+1);
    cout << b.size() << endl;
    REP(i, b.size()){
        cout << b[i];
        if(i == b.size()-1) cout << endl;
        else cout << " ";
    }
}