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>
#define rep(i, n) for (int i=0; i<(int)(n); i++)
#define all(v) v.begin(), v.end()
#define PRINT(v) for (auto x : (V)) cout <<x <<" " <<endl;
using namespace std;
using ll = long long;
using Graph = vector<vector<int>>;
const ll MOD = 1000000007;
const ll INF = 10000000000000000;
vector<int> x4 = {0, 1, 0, -1}, x8 = {0, 1, 1, 1, 0, -1, -1, -1};
vector<int> y4 = {1, 0, -1, 0}, y8 = {1, 1, 0, -1, -1, -1, 0, 1};
template<class T> struct edge { int from, to; T cost;};
template<class T> inline bool chmin(T& a, T b){if (a>b){a = b; return true;}return false;}
template<class T> inline bool chmax(T& a, T b){if (a<b){a = b; return true;}return false;}
template<class T> inline T powerM(T a,T b){if (b==0) return 1;
T tmp = powerM(a,b/2); if (b%2==0) return tmp*tmp%MOD; else return tmp*tmp%MOD*a%MOD; }
template<class T> inline T power(T a,T b,T m){ if (b==0) return 1;
  T tmp = power(a,b/2,m); if (b%2==0) return tmp*tmp%m; else return tmp*tmp%m*a%m; }
template<class T> inline T gcd(T a, T b){if (b==0) return a; return gcd(b, a%b);}
template<class T> inline T lcm(T a, T b){return a / gcd(a,b) * b;}
// ax+by=gcd(a,b)を解く
template<class T> inline T extgcd(T a,T b,T &x,T &y){if (b==0){x=1; y=0; return a;} T d=extgcd(b,a%b,y,x); y -= a/b*x; return d;}
void hey(){ cout <<"hey" <<endl; }


int N;
vector<int> A;
vector<bool> res;

void input() {
  cin >>N;
  A.assign(N, 0); rep(i, N) cin >>A[i];
  res.assign(N, false);
}

int main() {
  input();
  for (int i=N; i>0; i--){
    // 1-indexedに注意
    // iの倍数が書かれた箱に入っているボールの和がA[i-1]となるように、後ろから番号を決定していく
    if (i*2 > N){
      // iの倍数が自分自身しかいないなら、条件をそのまま自分に反映させる
      res[i-1] = A[i-1];
    }else{
      // iの倍数が自分以外にいるなら、自分以外のiの倍数の箱のボールの個数の和を計算する
      // 計算結果を2で割った余りがA[i-1]に一致するなら自分自身は0、一致しなければ自分自身は1とすればよい
      int sum = 0;
      for (int j=2*i; j<=N; j+=i) if (res[j-1]) sum++;
      sum %= 2;
      if (sum != A[i-1]) res[i-1] = true;
    }
  }
  int M = 0;
  rep(i, N) if (res[i]) M++;
  cout <<M <<endl;

  if (M > 0){
    rep(i, N) if (res[i]) cout <<i+1 <<" ";
    cout <<endl;
  }
}