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>
#pragma GCC optimize("Ofast")
using namespace std;
#define all(x) (x).begin(),(x).end()
#define rall(x) (x).rbegin(),(x).rend()
typedef long long ll;
#define rep(i,n) for(int i=0, i##_len=(n); i<i##_len; ++i)
#define REP(i,num,n) for(int i=num, i##_len=(n); i<i##_len; ++i)
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
#define print(x) (cout << (x) << endl)
#define pb push_back
#define mp make_pair
#define sz(x) int(x.size())
template<class T> inline void Yes(T condition){ if(condition) cout << "Yes" << endl; else cout << "No" << endl; }
#define fs first
#define sc second

const int INF = 1e9;
const ll LLINF = 1e16;
const ll MOD = 1e9+7;
template<class T> inline void add(T &a, T b){a = ((a+b) % MOD + MOD) % MOD;};

template<typename T>
vector<T> make_v(size_t a){return vector<T>(a);}
 
template<typename T,typename... Ts>
auto make_v(size_t a,Ts... ts){
  return vector<decltype(make_v<T>(ts...))>(a,make_v<T>(ts...));
}
 

// 指定した文字cが文字列に何文字入ってるか
ll stringcount(string s, char c) {
    return count(s.cbegin(), s.cend(), c);
}


// 割り算の切り上げ処理
template<typename T>
T rp(T a, T b) {
  return (a + b - 1) / b;
}

// 桁和
template<typename T>
T digsum(T n) {
    T res = 0;
    while(n > 0) {
        res += n%10;
        n /= 10;
    }
    return res;
}

// 回文判定
bool kai (string s) {
    string t{s};
    reverse(begin(t), end(t));
    return s == t;
}

using pint = pair<int, int>;
using pll = pair<ll, ll>;


int main(void)
{
    cout.tie(0);
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(20);
    ll n;
    cin >> n;
    vector<ll> a(n);
    rep(i, n) cin >> a[i];
    vector<ll> box(n);
    vector<ll> ans;
    for(int i = n - 1; i >= 0; --i){
        ll num = i + 1;
        ll sum = 0;
        for(int j = num + num; j < n; j += num){
            sum += box[j - 1];
        }
        if(sum % 2 != a[i]){
            box[i]++;
            ans.push_back(i + 1);
        }
    }
    print(ans.size());
    rep(i, sz(ans)){
        print(ans[i]);
    }

}