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;

#define REP(i,a,b) for(int i=a;i<(int)b;i++)
#define rep(i,n) REP(i,0,n)
#define all(c) (c).begin(), (c).end()
#define zero(a) memset(a, 0, sizeof a)
#define minus(a) memset(a, -1, sizeof a)
#define watch(a) { std::cout << #a << " = " << a << "\n"; }
template<class T1, class T2> inline bool minimize(T1 &a, T2 b) { return b < a && (a = b, 1); }
template<class T1, class T2> inline bool maximize(T1 &a, T2 b) { return a < b && (a = b, 1); }
template<class T, class V> istream& operator>> (istream& ist, pair<T, V>& p) { return ist >> p.first >> p.second; }
template<class T> ostream& operator<< (ostream& ost, pair<T, T>& p) { return ost << p.first << ", " << p.second; }
template<class T> istream& operator>> (istream& ist, vector<T>& vs) { for(auto& e: vs) ist >> e; return ist; }

typedef long long ll;
int const inf = INT_MAX / 2;

int main() {
  int N; cin >> N;
  vector<int> A(N + 1);
  rep(i, N) cin >> A[i + 1];
  vector<int> B;
  for (int i = N; i >= 1; i--) {
    int sum = 0;
    for (int j = 2 * i; j < N; j += i) {
      sum += A[j];
    }
    if (sum % 2 != A[i]) {
      B.push_back(i);
    }
  }
  cout << B.size() << "\n";
  if (B.size()) {
    rep(i, B.size()) {
      if (i) cout << " ";
      cout << B[i];
    }
    cout << "\n";
  }
}