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>
#include <boost/range/irange.hpp>

using namespace std;
using namespace boost;

main() {
    int64_t n;
    cin >> n;

    vector<int64_t> a(n);
    for (auto&& aa : a) {
        cin >> aa;
    }

    vector<int64_t> b(n, -1);
    for (auto i : irange(n - 1, -1L, -1L)) {
        int64_t s = 0;
        for (int64_t j = 2; (i + 1) * j <= n; ++j) {
            s += b[(i + 1) * j - 1];
        }

        b[i] = (s - a[i] + 2) % 2;
    }

    auto m = count_if(b.begin(), b.end(), [](int64_t bb) { return bb; });
    cout << m << endl;
    const auto* delim = "";
    for (auto i : irange(0L, n)) {
        if (b[i]) {
            cout << delim << i + 1;
            delim = " ";
        }
    }
    cout << "\n";
}