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;
using ll = long long;
using ull = unsigned long long;

#ifndef __MACRO_H__
#define __MACRO_H__

#define all(collection)  (collection).begin(), (collection).end()		// begin to end
#define rep(i, begin, end) for (ll i = begin; i < end; i++)				// repeat
#define repr(i, begin, end) for (ll i = begin; end < i; i--)			// repeat reverse

#endif

class ABC134D
{
public:
	ABC134D()
	{
		cin >> n;
		a.resize(n+1);
		rep(i, 1, n+1)
			cin >> a[i];
	}
	void Run(void)
	{
		vector<ll> ans(n + 1);


		repr(i, n, 0)
		{
			if (n < i * 2)
				ans[i] = a[i];
			else
				ans[i] = (SumAns(i, ans) % 2 == a[i]) ? 0 : 1;
		}

		cout << count(all(ans), 1) << endl;
		rep(i, 1, n + 1)
		{
			if (ans[i] == 1)
				cout << i << " ";
		}
	}
private:
	ll SumAns(ll s, vector<ll> ans)
	{
		ll count = 0;

		ll i = s * 2;
		while (n < i)
		{
			count += ans[i];
			i += s;
		}

		return count;
	}
	ll n;
	vector<ll> a;
};

int main(void)
{
	ABC134D abc134d;
	abc134d.Run();

	return 0;
}