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;
typedef long long ll;
#define rep(i, a, b) for (ll i = a; i < b; i++)
#define all(v) v.begin(), v.end()
const int MOD = 1e9 + 7;
const int INF = 1e9;
//'A' = 65, 'Z' = 90, 'a' = 97, 'z' = 122

int main() {
    ll n;
    cin >> n;
    ll a[n];
    rep(i, 0, n) {
        cin >> a[i];
    }

    if (n == 1) {
        if (a[0]) {
            cout << 1 << endl;
            cout << 1 << endl;
        } else {
            cout << 0 << endl;
        }
        return 0;
    }

    ll num[n + 1] = {};
    rep(i, 1, n + 1) {
        num[i] = n / i;
    }

    bool flag[n + 1] = {};
    ll x = (n + 1) / 2;
    for (ll i = n; i >= x; i--) {
        if (a[i - 1] == 1) flag[i] = true;
    }

    for (ll i = x - 1; i >= 1; i --) {
        ll sum = 0;
        for (ll j = i; j <= n; j += i) {
            sum += (ll)flag[j];
        }
        if (sum % 2 != a[i - 1]) flag[i] = true;
    }
    ll m = 0;
    rep(i, 1, n + 1) {
        if (flag[i]) m++;
    }

    cout << m << endl;
    rep(i, 1, n + 1) {
        if (flag[i]) cout << i << " ";
    }
    cout << endl;

    return 0;
}