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;

#define ll long long
#define fw(p) for(int w=0;w<(p);w++)
#define fx(p) for(int x=0;x<(p);x++)
#define fy(p) for(int y=0;y<(p);y++)
#define fz(p) for(int z=0;z<(p);z++)
#define fyg(p,g) for(int y=(g);y<(p);y++)
#define fzg(p,g) for(int z=(g);z<(p);z++)
#define ce(d) cout<<d<<endl;
#define vecp(p) int aa;cin>>aa;(p).push_back(aa);
#define vecpl(p) long long aa;cin>>aa;(p).push_back(aa);
#define vecps(p) string aa;cin>>aa;(p).push_back(aa);
#define vecp2(p) cin>>aa;(p).push_back(aa);
#define vecpl2(p) long long a	b;cin>>ab;(p).push_back(ab);
#define vecps2(p) string ab;cin>>ab;(p).push_back(ab);
#define set0(k,n) for(int nn=0;nn<(n);nn++){ (k).push_back(0); }
#define sorts(c) sort((c).begin(),(c).end());
#define reverses(c) reverse((c).begin(),(c).end());
#define vec(b) vector<int> (b);
#define vecl(b) vector<long long> (b);
#define vecs(b) vector<string> (b);
#define vecsize(b,size) vector<int> (b)((size));
#define pb(b,a) (b).push_back((a));
#define doublece(a,b) cout<<(a)<<' '<<(b)<<endl;
#define pairs(s) vector<pair<int,int>> (s);
#define pairsl(s) vector<pair<ll,ll>> (s);
#define pairss(s) vector<pair<string,string>> (s);
#define pairsp(s) int aa,bb;cin>>aa>>bb;(s).push_back(make_pair(aa,bb));
#define pairspl(s) int aa,bb;cin>>aa>>bb;(s).push_back(make_pair(aa,bb));
#define pairsps(s) int aa,bb;cin>>aa>>bb;(s).push_back(make_pair(aa,bb));
#define pairsREV(s) (s).push_back(make_pair(bb,aa));
#define pairslREV(s) (s).push_back(make_pair(bb,aa));
#define pairssREV(s) (s).push_back(make_pair(bb,aa));
#define MOD 1000000007
int ANS[210000];
int main()
{

	int N;
	cin >> N;
	vec(A)
		pb(A,0)
	fx(N) {
		vecp(A)
	}

	int M = 0;
	for (int x = N / 2 + 1; x <= N;x++) {
		if (A.at(x)==1) {
			ANS[x] = 1;
			M++;
		}
	}

	for (int x = N / 2; x > 0;x--) {
		int i = 2;
		int sum = 0;
		while (x*i<=N) {
			sum += ANS[x*i];
			i++;
		}
		ANS[x] = sum % 2;
		if (ANS[x]!=A.at(x)) {
			ANS[x] = 1;
		}
		M += ANS[x];
	}

	if (ANS[1]==A.at(1)) {
		ce(M)
		fx(N) {
			if (ANS[x + 1]==1) {
				cout << x+1<<' ';
			}
		}
	}
	else {
		fx(N) {
				cout << ANS[x+1];

		}
		ce(-1)
	}

    return 0;
}