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: Wrong Answer)

import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.HashSet;
import java.util.List;
import java.util.Map;
import java.util.Map.Entry;
import java.util.Scanner;
import java.util.Set;
 
 
public class Main {

	static long MOD = 1_000_000_007;
	static int INF = 1_000_000_007;
	//二項係数関連ここから-----
	static boolean isPreprocessing = false;	//二項係数を計算するための前処理を行っているかどうかのフラグ
	static int MAX = 260;				//MAXまで階乗やそれらの逆元のテーブルを保持する。
	static long[] fac;	//fac[i] := i! % MOD
	static long[] inv;	//inv[i] := mod. MODにおける i の逆元
	static long[] finv;	//finv[i] := mod. MODにおける i! の逆元(invの累積"積")
	//二項係数関連おわり------
	
	public static void main(String[] args) throws Exception {
		Scanner sc = new Scanner(System.in);
		//int N = Integer.parseInt(sc.next());
		//long L = Long.parseLong(sc.next());
		//char[] A = sc.next().toCharArray();
		
		int N = Integer.parseInt(sc.next());
		int[] A = new int[N];
		int[] B = new int[N];
		int half = N/2 ;
		for(int i=0; i<N;i++){
			A[i]=Integer.parseInt(sc.next());
			if(i>=half)B[i] = A[i];
		}
		if(N==1){
			if(A[0]==1){
				System.out.println(1);
				System.out.println(A[0]);
			}else{
				System.out.println(0);
			}
		}else{
			for(int i=half; i>0; i--){
				int tmp = i;
				int ball = 0;
				tmp += i;
				while(tmp<=N){
					ball += A[tmp-1];
					tmp += i;
				}
				B[i-1] = A[i-1]^(ball%2);
			}
			
			int M = 0;
			for(int b : B){
				M += b;
			}
			System.out.println(M);
			for(int i=0; i<N; i++){
				if(B[i]==1)System.out.print((i+1) + " ");
			}
		}
		

		

		
		
	}
	
	//aとbの最小公倍数を返す
	static long lcm (long a, long b) {
		long temp;
		long c = a;
		c *= b;
		while((temp = a%b)!=0) {
			a = b;
			b = temp;
		}
		return (c/b);
	}
	
	//二項係数用の前処理
	static void COMinit(){
		//二項係数演算用前処理
		fac = new long[MAX+1];
		fac[0] = 1L;
		fac[1] = 1L;
		inv = new long[MAX+1];
		inv[1] = 1L;
		finv = new long[MAX+1];
		finv[0] = 1L;
		finv[1] = 1L;
		for(int i=2;i<MAX;i++){
			fac[i] = fac[i-1] * i % MOD;
			inv[i] = MOD - inv[(int)(MOD%i)] * (MOD/i) % MOD;
			finv[i] = finv[i-1] * inv[i] % MOD;
		}
		isPreprocessing = true;
	}
	
	//二項係数i_C_jをMODで割った余りを返す
	static long COM(int i, int j){
		if(i<j)return 0;
		if(i<0 || j<0)return 0;
		if(!isPreprocessing)COMinit();
		return fac[i] * (finv[j] * finv[i-j] % MOD) % MOD;
	}
	
}