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 <iostream>
#include <vector>
#include <algorithm>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <climits>
#include <numeric>

using namespace std;

#define REP(i,n) for(int i=0, i##_len=(n); i<i##_len; ++i)

typedef long long LL;
typedef vector<int> VI;

int main() {
  int n;
  cin >> n;
  vector<int> a(n+1);
  for (int i = 1; i <= n; i++)
    cin >> a[i];
  vector<int> dp(n+1);
  for (int i = n; i >= 1; i--) {
    if (2*i > n) {
      dp[i] = a[i];
    } else {
      int k = 2;
      while (k*i <= n) {
        dp[i] ^= dp[k*i];
        k++;
      }
      dp[i] ^= a[i];
    }
  }
  int m = accumulate(dp.begin(), dp.end(), 0);
  cout << m << endl;
  for (int i = 1; i <= n; i++) {
    if (dp[i])
      cout << i << endl;
  }
  return 0;
}