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;

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

	set<int64_t> st;

	for (int64_t i = n - 1; n < 2 * (i + 1); i--)
	{
		if (as[i])
		{
			st.insert(i + 1);
		}
	}

	for (int64_t i = 0; i < n / 2; i++)
	{
		int64_t cnt = 0;
		for (int64_t j = 2; (i + 1) * j <= n; j++)
		{
			if (st.find(j) != st.end())
			{
				cnt++;
			}
		}
		if ((cnt % 2) != as[i])
		{
			st.insert(i + 1);
		}
	}

	cout << st.size() << endl;
	for (auto &a : st)
	{
		cout << a << ' ';
	}
	cout << endl;

	return 0;
}