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<iostream>
#include<string>
#include<algorithm>
#include<vector>
#include<iomanip>
#include<math.h>
#include<complex>
#include<queue>
#include<deque>
#include<stack>
#include<map>
#include<set>
#include<bitset>
#include<functional>
#include<assert.h>
#include<numeric>
#include<stdio.h>
#include <cstdint>
#include <stdlib.h> 
#include <time.h>
using namespace std;
#define REP(i,m,n) for(int i=(int)(m) ; i < (int) (n) ; ++i )
#define rep(i,n) REP(i,0,n)
#define rrep(i,n) REP(i,1,n+1)
#define all(a) a.begin(), a.end()
#define fi first
#define se second
#define RNG(x, a, n) &((x)[a]), &((x)[n])
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
typedef long long ll;
typedef pair<ll, ll> Pll;
typedef pair<int, int> PII;
typedef vector<ll> Vl;
typedef vector<int> VI;
typedef tuple<int, int, int> TT;
#define chmin(x,y) x = min(x,y)
#define chmax(x,y) x = max(x,y)
#define rall(v) v.rbegin(), v.rend()
#define dmp(x,y) make_pair(x,y)
#define dmt(x, y, z) make_tuple(x, y, z)
#define pb(x) push_back(x)
#define pf(x) push_front(x)
const int MAX = 2000000;
const int inf = 1000000007;
const ll mod = 1000000007;
const ll longinf = 1LL << 60;
const long double PI = (acos(-1));
const long double EPS = 0.0000000001;

int dx[4] = { 1, 0, -1, 0 };
int dy[4] = { 0, 1, 0, -1 };



int m, b[200005];
int tmp[200005];
VI vec;

int main() {

	cout << -1 << endl;

	return 0;
}