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 rep(i,n) for(int i=0;i<(n);++i)
#define rrep(i,n) for(int i=1;i<(n);++i)
#define all(a) (a).begin(),(a).end()
#define rall(a) (a).rbegin(),(a).rend()
using namespace std;
typedef long long ll;
typedef pair<int, int> P;
const int inf = 0x7ffffff;
const int mod = 1000000007;

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	int n;
	cin >> n;
	vector<int> a(n);
	rep(i, n) cin >> a[i];
	vector<int> p(n);
	for (int i = n - 1; i >= 0; --i) {
		int tmp = 0;
		for (int j = i + (i + 1); j < n; j += (i + 1)) {
			tmp += p[j];
		}
		if (tmp % 2 != a[i]) {
			p[i] = 1;
		}
	}
	vector<int> res;
	int ans = 0;
	rep(i, n) {
		if (p[i]) {
			ans++;
			res.emplace_back(i + 1);
		}
	}
	cout << ans << endl;
	for (int c : res) {
		if (c == res.back()) {
			cout << c << endl;
		}
		else {
			cout << c << ' ';
		}
	}
	return 0;
}