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 ld double
#define pii pair<ll,ll>
#define vl vector<ll>
#define vb vector<bool>
#define vs vector<string>
#define vp vector<pii>
#define pb push_back
#define mp make_pair
#define sc(a) scanf("%lld",&a);
#define sd(a) scanf("%lf",&a);
#define test() ll tt;cin>>tt;for(ll testcase=1;testcase<=tt;testcase++)
#define all(v) v.begin(),v.end()
#define vivek_ghosh ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define MAX 1000055
#define mod 1000000007
#define ff first
#define ss second
const long long int INF=0x3f3f3f3f3f3f;

ll power(ll x,ll n,ll M)
{
    if(n==0)
        return 1;
    else if(n%2 == 0)
        return power((x*x)%M,n/2,M);
    else
        return (x*power((x*x)%M,(n-1)/2,M))%M;
}

ll modinv(ll a,ll m)
{
    ll g =__gcd(a, m);
    ll res=power(a, m-2, m);
    return res;
}

vb prime(MAX,true);
vl prm;
void sieve()
{
	prime[0] = prime[1] = false;
	for(ll p=2;p*p<MAX;p++)
	{
		if(prime[p])
		{
			for(ll i=p*p;i<MAX;i+=p)
				prime[i] = false;
		}
	}
	for(ll p=2;p<MAX;p++)
	{
		if(prime[p])
			prm.pb(p);
	}
}

ll n,m,t,i,j,k,q,x,y,ans=0,cnt=0,sum=0,mx=-1,mn=INT_MAX;
string s;

int main()
{
    vivek_ghosh;

    cin>>n;
    
    vl a(n+1,0);
    
    for(i=1;i<=n;i++)
        cin>>a[i];
        
    ll id = (n/2);
    
    while(id>0)
    {
        cnt = 0;
        for(ll i=id+id;i<=n;i+=id)
        {
            cnt+=a[i];
            cnt%=2;
        }
        if(cnt == 1)
        {
            ans++;
            a[i] = 1;
        }
        id--;
    }
    
    cout<<ans<<endl;
    for(i=1;i<=n;i++)
    {
        if(a[i])
            cout<<1<<" ";
    }
    
    return 0;
}