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 in std::cin
#define out std::cout
#define rep(i,N) for(LL i=0;i<N;++i)
typedef long long int LL;

int main()
{
	LL N;
	in >> N;
	std::vector<LL>a(N + 1);
	rep(i, N) in >> a[i + 1];

	std::vector<LL>ans(N + 1);
	for (LL i = N; i >= 1; --i)
	{
		if (i * 2 > N) ans[i] = a[i];
		else
		{
			LL sum = 0;
			for (LL j = i * 2; j <= N; j += i) sum += ans[j];
			sum %= 2;
			if (sum != a[i]) ans[i] = 1;
		}
	}

	std::vector<LL>b;
	for (LL i = 1; i <= N; ++i)
	{
		if (ans[i] == 1) b.push_back(i);
	}
	out << b.size() << std::endl;
	rep(i, b.size()) out << b[i] << std::endl;
}