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>
using namespace std;
	
#define int long long
#define pb push_back
#define ub upper_bound
#define lb upper_bound
#define mp make_pair
#define F first
#define S second
#define FOR(i,a,b) for(int (i)=(a);(i)<(b);(i)++)
#define REP(i,n) FOR(i,0,n)
#define ALL(a) (a).begin(),(a).end()
#define RALL(a) (a).rbegin(),(a).rend()
#define SORT(c) sort((c).begin(),(c).end())
#define ve vector
#define vi vector<int>
#define vp vector<pair<int,int>>
#define vvi vector<vector<int>>
	
using ll = long long;
const ll INF = LLONG_MAX - 100;
ll mod = 1e9 + 7;
int dx[] = {-1,0,1,0}, dy[] = {0,1,0,-1};
vector<ll> prime;
	
class fact {
public:
	int fmod = 1e9+7;
	vector<int> fac, inv;
	fact (int n, int Mod = 1e9+7) {
		fmod = Mod;
		fac = vector<int>(n,0);
		inv = vector<int>(n,0);
		fac[0] = 1; for (int i = 1; i < n; i++) fac[i] = fac[i-1] * i % fmod;
		for (int i = 0;i < n;i++) inv[i] = fact::POW(fac[i],fmod-2);
	}
	ll nCr(ll n, ll r) {return (fac[n] * inv[r] % fmod) * inv[n-r] % fmod;}
	ll POW(ll a, ll b) {ll c = 1; while (b > 0) {if (b & 1) {c = a * c%fmod;}a = a * a%fmod; b >>= 1;}return c;}
};
	
template <class T = ll> T in() {T x; cin >> x; return (x);}
void DEBUG(vector<int> a) {for(int i=0;i<a.size();i++)cout<<a[i]<<" ";cout<<endl;}
void EMP(int x) {cout<<"!!!"<<x<<"!!!"<<endl;}
ll GCD(ll a, ll b) {ll c; while (b != 0) {c = a % b; a = b; b = c;}return a;}
ll LCM(ll a, ll b) {return a * b / GCD(a, b);}
ll POW(ll a, ll b) {ll c = 1; while (b > 0) {if (b & 1) {c = a * c%mod;}a = a * a%mod; b >>= 1;}return c;}
void PRI(ll n) {bool a[n + 1LL]; for (int i = 0; i < n + 1LL; i++) {a[i] = 1LL;}for (int i = 2; i < n + 1LL; i++) {if (a[i]) {prime.pb(i); ll b = i; while (b <= n) {a[b] = 0; b += i;}}}}
template <typename T> T chmin(T& a, T b) {if(a>b)a=b;return a;}
template <typename T> T chmax(T& a, T b) {if(a<b)a=b;return b;}
bool isSqrt(ll a) {return pow(sqrt(a),2) == a ? 1 : 0;}
void YesNo(bool a) {if (a) cout << "Yes"; else cout << "No"; cout << endl;}
void yesno(bool a) {if (a) cout << "yes"; else cout << "no"; cout << endl;}
void YESNO(bool a) {if (a) cout << "YES"; else cout << "NO"; cout << endl;}
	
bool solve() {
	int n; cin >> n;
	int a[n]; REP (i,n) cin >> a[i];
	bool flag = true; int b[n];
	int border; 
	for (int i = n/2+1;i <= n;i++) {
		b[i-1] = a[i-1];
	}
	for (int i = n/2;i >= 1;i--) {
		int p = i * 2;
		int tmp = 0;
		while (p <= n) {
			tmp += b[p-1];
			p += i;
		}
		if (tmp%2 == a[i-1]) b[i-1] = 0;
		else b[i-1] = 1;
	}
	int cnt = 0;
	REP (i,n) {
		if (b[i]) cnt++;
	}
	cout << cnt << endl;
	REP (i,n) {
		if (b[i]) cout << i+1 << endl;
	}
}
	
signed main() {
	// cin.tie(0);
	// ios::sync_with_stdio(false);
	solve();
}