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>
        #define ll long long
        #define rep(i,a,b) for(i=a;i<b;i++)
        #define pb push_back
        #define all(v) v.begin(),v.end()
        #define F first
        #define S second
        #define vi vector < int >
        #define vii vector < long long int>
        #define CLR(a) memset((a),0,sizeof(a))
        #define mp make_pair
        #define hell 1000000007

        using namespace std;


        void solve()
        {
           ll int i,j=0,r=200,p,q,k=0,d,y=0,h,flag=0,n,m=0,W;
            cin>>n;
            bool v[n+1],vis[n+1];
            memset(vis,false,sizeof(vis));
            vector < ll >as;
            rep(i,1,n+1) cin>>v[i];

             for(i=n;i>n/2;i--)
             {
                 if(v[i]) as.pb(i),vis[i]=true;
             }
             for(i=n/2;i>0;i--)
             {
                y=0;
                 for(j=2*i;j<=n;j+=i)
                 {
                     if(vis[j]) y++;
                 }
                 if(v[i]) y++;
                 if(y%2==v[i]%2){if(v[i]) as.pb(i);}
                 else {cout<<-1; return;}
             }

              cout<<as.size()<<endl;
              rep(i,0,as.size()) cout<<as[i]<<" ";


        }


        int main()
        {
            ios::sync_with_stdio(false);
            cin.tie(0);
            cout.tie(0);
            int t=1;
            //cin>>t;
            while(t--)
            solve();
        }

           /*

      void bfs( int i , int j)
           {
               vis[i][j]= true;
             queue < int > que;
             que.push(k);
             while(!que.empty())
             {
                int x= que.front();
                que.pop();
                 for ( int i=1;i<=n;i++)
                 {
                     if((!g[1][n] and g[x][i] ) or (g[1][n] and ! g[x][i]))
                     {
                       if(!vis[i])
                       {
                           que.push(i);
                           level[i]=level[x]+1;
                           vis[i]= true;
                       }
                     }
                 }
             }
           }

          ll int por(ll int a,ll int b)
             {
                ll int k=1;
                while(b>0)
                {
                    if(b%2) k=k*a,k%=hell;
                    b/=2;
                    a*=a; a%=hell;
                }
                return k;
             } */