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 mod 1000000007
#define pi 3.14159265358979323846264
#define pii pair <int, int>
#define pb push_back
#define mp make_pair
#define vi vector <int>
#define vl vector <long long>
#define vpii vector <pair <int, int> >
#define ll long long
#define fi first
#define se second
#define FOR(i, x, y) for (int i=x; i<=y; i++)
#define FORI(i, y, x) for (int i=y; i>=x; i--)
#define debug printf("OK\n")
int n, a[200005], c[200005], d[200005], e[200005];
vector <int> v;
int main() {
    cin >> n;
    FOR(i, 1, n) cin >> a[i];
    int k = n/2 + n%2;
    FOR(i, k, n) {
        c[i] = a[i];
        if (a[i] == 1) v.pb(i);
    }
    for (int i=k-1; i>=1; i--) {
        int cnt = 0;
        for (int j=i+i; j<=n; j+=i) cnt += c[j];
        //printf("%d %d\n", i, cnt);
        cnt %= 2;
        if (cnt != a[i]) {
            c[i]++;
            v.pb(i);
        }
    }
    int size = v.size();
    FOR(i, 0, size-1) {
        d[v[i]] = 1;
    }
    for (int i=1; i<=n; i++) {
        for (int j=i; j<=n; j+=i) e[i] += d[j];
    }
    FOR(i, 1, n) {
        //cout << d[i] << " ";
        if (e[i] % 2 != a[i]) {
            printf("-1\n");
             return 0;
        }
    }
    cout << size << endl;
    FOR(i, 0, size-1) printf("%d%c", v[i], i==size-1? '\n' : ' ');
    return 0;
}