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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#define all(v) v.begin(), v.end()
#define rall(v) v.rbegin(), v.rend()
#define rep(i,n) for (int i=0;i<(int)(n);i++)
#define codefor int test;scanf("%d",&test);while(test--)
#define INT(...) int __VA_ARGS__;in(__VA_ARGS__)
#define LL(...) ll __VA_ARGS__;in(__VA_ARGS__)
#define yn(ans) if(ans)printf("Yes\n");else printf("No\n")
#define YN(ans) if(ans)printf("YES\n");else printf("NO\n")
#define vector1d(type,name,...) vector<type>name(__VA_ARGS__)
#define vector2d(type,name,h,...) vector<vector<type>>name(h,vector<type>(__VA_ARGS__))
#define umap unordered_map
#define uset unordered_set
using namespace std;
using ll = long long;
const int MOD=1000000007;
//入力系
void scan(int& a){scanf("%d",&a);}
void scan(long long& a){scanf("%lld",&a);}
template<class T> void scan(T& a){cin>>a;}
template<class T> void scan(vector<T>& vec){for(auto&& it:vec)scan(it);}
void in(){}
template <class Head, class... Tail> void in(Head& head, Tail&... tail){scan(head);in(tail...);}
//出力系
void print(const int& a){printf("%d",a);}
void print(const long long& a){printf("%lld",a);}
void print(const double& a){printf("%.15lf",a);}
template<class T> void print(const T& a){cout<<a;}
template<class T> void print(const vector<T>& vec){if(vec.empty())return;print(vec[0]);for(auto it=vec.begin();++it!= vec.end();){putchar(' ');print(*it);}}
void out(){putchar('\n');}
template<class T> void out(const T& t){print(t);putchar('\n');}
template <class Head, class... Tail> void out(const Head& head,const Tail&... tail){print(head);putchar(' ');out(tail...);}
//デバッグ系
template<class T> void dprint(const T& a){cerr<<a;}
template<class T> void dprint(const vector<T>& vec){if(vec.empty())return;cerr<<vec[0];for(auto it=vec.begin();++it!= vec.end();){cerr<<" "<<*it;}}
void debug(){cerr<<endl;}
template<class T> void debug(const T& t){dprint(t);cerr<<endl;}
template <class Head, class... Tail> void debug(const Head& head, const Tail&... tail){dprint(head);cerr<<" ";debug(tail...);}
ll intpow(ll a, ll b){ ll ans = 1; while(b){ if(b & 1) ans *= a; a *= a; b /= 2; } return ans; }
ll modpow(ll a, ll b, ll p){ ll ans = 1; while(b){ if(b & 1) (ans *= a) %= p; (a *= a) %= p; b /= 2; } return ans; }
ll updivide(ll a,ll b){if(b==1) return a;else if(a%b==0) return a/b;else return (a/b)+1;}

int main(){
    INT(n);
    vector<int> vec(n+1,0);
    vector<int> ans(n+1,0);
    ll sum=0;
    for(int i=1;i<n+1;i++)in(vec[i]);
    for(int i=n;i>=1;i--){
        for(int j=2*i;j<=n;j+=i){
            ans[i]+=ans[j];
        }
        if(sum%2!=vec[i])ans[i]+=1;
    }
    int zero=count(all(ans),0);
    out(n+1-zero);
    if(n+1-zero){
        for(int i=1;i<n+1;i++){
            if(ans[i]!=0)printf("%d ",ans[i]);
        }
        cout<<endl;
    }
}