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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>

using namespace std;
 
typedef long long ll;
typedef long double ld;
typedef pair<int, int> Pii;
typedef pair<int, ll> Pil;
typedef pair<ll, ll> Pll;
typedef pair<ll, int> Pli;
typedef vector < vector<ll> > Mat;

#define fi first
#define se second

const ll MOD = 1e9 + 7;
const ll MOD2 = 998244353;
const ll MOD3 = 1812447359;
const ll INF = 1ll << 62;
const double PI = 2 * asin(1);

void yes() {printf("yes\n");}
void no() {printf("no\n");}
void Yes() {printf("Yes\n");}
void No() {printf("No\n");}
void YES() {printf("YES\n");}
void NO() {printf("NO\n");}

int N, A[int(2e5+5)], ans[int(2e5+5)];
vector <int> res;

int main(){
  cin >> N;
  for (int i = 1; i <= N; i++) cin >> A[i];

  for (int i = N; i >= 1; i--){
    if (ans[i] % 2 == A[i]) continue;
    for (int j = 1; j * j <= i; j++){
      if (i % j != 0) continue;
      ans[j]++;
      if (j * j != i) ans[i/j]++;
    }
  }

  for (int i = 1; i <= N; i++){
    if (ans[i] % 2 != A[i]){
      cout << -1 << endl;
      return 0;
    }
  }


  for (int i = 1; i <= N; i++){
    if (ans[i] % 2 != 0) res.push_back(i);
  }
  int M = res.size();
  cout << M << endl;
  for (int i = 0; i < M; i++){
    cout << res[i];
    if (i < M - 1) cout << " ";
    else cout << endl;
  }

  return 0;
}