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)

#pragma GCC optimize ("Ofast")
#include<bits/stdc++.h>
#define pb push_back
#define __V vector
#define all(x) x.begin(), x.end()
#define oit ostream_iterator
#define mod 1000000007ll
using namespace std;
void doin() {
	cin.tie();
	cout.tie();
	ios::sync_with_stdio(0);
#ifndef ONLINE_JUDGE
//	freopen("input.txt", "r", stdin);
//	freopen("output.txt", "w", stdout);
#endif
}
template<typename X, typename Y>
istream& operator>>(istream& in, pair<X, Y> &a) {
	in >> a.first >> a.second;
	return in;
}
template<typename T>
void getv(T& i) {
	cin >> i;
}
template<typename T, typename ... Ns>
void getv(vector<T>& v, int n, Ns ... ns) {
	v.resize(n);
	for (auto& i : v)
		getv(i, ns...);
}
template<typename T>
void getv1(T& i) {
	cin >> i;
}
template<typename T, typename ... Ns>
void getv1(vector<T>& v, int n, Ns ... ns) {
	v.resize(n + 1);
	for (int i = 1; i <= n; i++)
		getv(v[i], ns...);
}
#ifdef _WIN32
#define getchar_unlocked() _getchar_nolock()
#define _CRT_DISABLE_PERFCRIT_LOCKS
#endif
inline void getch(char &x) {
	while (x = getchar_unlocked(), x < 33) {
		;
	}
}
inline void getstr(string &str) {
	str.clear();
	char cur;
	while (cur = getchar_unlocked(), cur < 33) {
		;
	}
	while (cur > 32) {
		str += cur;
		cur = getchar_unlocked();
	}
}
template<typename T> inline bool sc(T &num) {
	bool neg = 0;
	int c;
	num = 0;
	while (c = getchar_unlocked(), c < 33) {
		if (c == EOF)
			return false;
	}
	if (c == '-') {
		neg = 1;
		c = getchar_unlocked();
	}
	for (; c > 47; c = getchar_unlocked())
		num = num * 10 + c - 48;
	if (neg)
		num *= -1;
	return true;
}
typedef unsigned long long ull;
typedef long long ll;
typedef float ld;
typedef ll _I;
typedef pair<_I, _I> pi;
typedef pair<ld, ld> pd;
typedef map<_I, _I> mii;
typedef __V <_I> vi;
typedef __V <char> vc;
typedef __V <ld> vd;
typedef __V <vd> vvd;
typedef __V <pi> vpi;
typedef __V <__V<_I>> vvi;
typedef __V <__V<char>> vvc;
typedef __V <__V<pi>> vvpi;
using AntonTsypko = void;
using arsijo = AntonTsypko;
using god = arsijo;
uniform_real_distribution<double> double_dist(0, 1);
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

int main() {
	doin();
	ll n, pm, c;
	vi v;
	cin >> n;
	getv1(v, n);
	for (auto& i : v)
		i &= 1;
	vi a(n+1, 0), r;
	for(int i = 1; i <= n; i++) {
		if(a[i] != v[i]) {
			r.pb(i);
			for(int j = i; j <= n; j+=i)
				a[j]^=1;
		}
	}
	cout << r.size() << "\n";
	for(auto i : r) cout << i << " ";
	return 0;
}