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 <bits/stdc++.h>

using namespace std;

using ll = long long;

int main(){
  ll n;
  cin >> n;
  cin.ignore();
  vector <ll> a;
  ll tmp = 0;
  bool allzero = true;
  for (ll i = 0; i < n; i++){
    cin >> tmp;
    if(tmp != 0){
      allzero = false;
    }
    a.push_back(tmp);
  }
  ll b[n] = {0};
  for(ll i = 0; i < n/2; i++){
    b[i] = 0;
  }
  for(ll i = n/2; i < n; i++){
    b[i] = a[i];
  }

  if(allzero == true){
    cout << 0 << endl;
    return 0;
  }

  // 後ろからあてはめていく
  ll sufi, sufj;
  ll sum;
  for (ll i = n/2; i >= 1; i--){
    sufi = i - 1;
    sum = 0;
    for(ll j = i * 2; j <= n; j += i){
      sufj = j - 1;
      sum += b[sufj];
    }
    sum = sum % 2;
    if(a[sufi] != sum){
      //cout << i << endl;
      b[sufi] = a[sufi] + 1;
      b[sufi] %= 2;
    } else {
      b[sufi] = a[sufi];
    }
  }

  // ここから変数の意味が変わる
  sum = 0;
  for (ll i = 0; i < n; i++){
    if(b[i] == 1) ++sum;
  }

  cout << sum << endl;
  
  for (ll i = 0; i < n; i++){
    if(b[i] == 1){
      cout << (i + 1) << " " << flush;
    }
  }
  cout << endl;
  return 0;
}