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: Accepted)

#include<bits/stdc++.h>
#define forr(i, a, b) for(int i = (a); i < (b); i++)
#define rep(i,n) for(int i = 0; i < (n); i++)
#define rrep(i,n) for(int i = 1; i <= (n); i++)
#define ALL(a) (a.begin()),(a.end())
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
typedef pair<ll, ll> LP;
const ll LINF = 1LL<<60;
const int INF = 1001001001;
const int MOD = 1000000007;

/* --------------------------------------------------- */

int main() {
    int n;
    cin >> n;
    vector<int> a(n);
    rep(i, n) cin >> a[i];
    vector<int> ans(n + 1, 0);
    for(int i = n; i > 0; i--) {
        if(i > n / 2) {
            if(a[i - 1] == 1) ans[i] = 1;
        } else {
            int cnt = 0;
            for(int j = i; j <= n; j += i) cnt += ans[j];
            cnt %= 2;
            if(a[i - 1] == 1 && cnt == 0) ans[i] = 1;
            else if(a[i - 1] == 0 && cnt == 1) ans[i] = 1;
        }
    }
    int m = 0;
    rrep(i, n) m += ans[i];
    cout << m << endl;
    rrep(i, n) {
        if(ans[i] == 1) cout << i << " ";
    } 
    cout << endl;

    return 0;
}