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 Java8 (OpenJDK 1.8.0) (judge verdict: Time Limit Exceeded)

import java.util.Scanner;

public class Main {

    void solve() {
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        int[] a = new int[N + 1];
        for (int i = 1; i <= N; i++) {
            a[i] = sc.nextInt();
        }
        int[] b = new int[N + 1];

        int allCount = 0;
        // 少なくとも後半半分は a[i] がそのまま b[i] になる。
        for (int i = N; i >= 1; i--) {
            if (i >= (N/2 + 1)) {
                b[i] = a[i];
                if (b[i] == 1) {
                    allCount++;
                }
            }
            else {
                int count = 0;
                // ★ここを高速化したいが、どうしたらいいのか・・。
                if (i == 1) {
                    count = allCount;
                }
                else {
                    for (int j = i + 1; j <= N; j++) {
                        if (j % i == 0) {
                            if (b[j] == 1) {
                                count++;
                            }
                        }
                    }
                }

                // count と a[i] に合わせて、b[i] を決める。
                if (count % 2 == 0) {
                    if (a[i] == 0) {
                        b[i] = 0;
                    }
                    else { // a[i] == 1
                        b[i] = 1;
                        allCount++;
                    }
                }
                else { // count % 2 == 1
                    if (a[i] == 0) {
                        b[i] = 1;
                        allCount++;
                    }
                    else { // a[i] == 1
                        b[i] = 0;
                    }
                }
            }
        }
        System.out.println(allCount);
        for (int i = 1; i <= N; i++) {
            if (b[i] == 1) {
                System.out.print(i + " ");
            }
        }
    }
    // Main
    public static void main(String[] args) {
        new Main().solve();
    }
}