When Masha came to math classes today, she saw two integer sequences of length n - 1 on the blackboard. Let's denote the elements of the first sequence as a_i (0 ≤ a_i ≤ 3), and the elements of the second sequence as b_i (0 ≤ b_i ≤ 3).

Masha became interested if or not there is an integer sequence of length n, which elements we will denote as t_i (0 ≤ t_i ≤ 3), so that for every i (1 ≤ i ≤ n - 1) the following is true: 

  * a_i = t_i | t_{i + 1} (where | denotes the [bitwise OR operation](https://en.wikipedia.org/wiki/Bitwise_operation#OR)) and 
  * b_i = t_i \& t_{i + 1} (where \& denotes the [bitwise AND operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND)). 



The question appeared to be too difficult for Masha, so now she asked you to check whether such a sequence t_i of length n exists. If it exists, find such a sequence. If there are multiple such sequences, find any of them.

Input

The first line contains a single integer n (2 ≤ n ≤ 10^5) — the length of the sequence t_i. 

The second line contains n - 1 integers a_1, a_2, …, a_{n-1} (0 ≤ a_i ≤ 3) — the first sequence on the blackboard.

The third line contains n - 1 integers b_1, b_2, …, b_{n-1} (0 ≤ b_i ≤ 3) — the second sequence on the blackboard.

Output

In the first line print "YES" (without quotes), if there is a sequence t_i that satisfies the conditions from the statements, and "NO" (without quotes), if there is no such sequence.

If there is such a sequence, on the second line print n integers t_1, t_2, …, t_n (0 ≤ t_i ≤ 3) — the sequence that satisfies the statements conditions.

If there are multiple answers, print any of them.

Examples

Input

4
3 3 2
1 2 0


Output

YES
1 3 2 0 

Input

3
1 3
3 2


Output

NO

Note

In the first example it's easy to see that the sequence from output satisfies the given conditions: 

  * t_1 | t_2 = (01_2) | (11_2) = (11_2) = 3 = a_1 and t_1 \& t_2 = (01_2) \& (11_2) = (01_2) = 1 = b_1; 
  * t_2 | t_3 = (11_2) | (10_2) = (11_2) = 3 = a_2 and t_2 \& t_3 = (11_2) \& (10_2) = (10_2) = 2 = b_2; 
  * t_3 | t_4 = (10_2) | (00_2) = (10_2) = 2 = a_3 and t_3 \& t_4 = (10_2) \& (00_2) = (00_2) = 0 = b_3. 



In the second example there is no such sequence.
import java.lang.reflect.Array;
import java.util.Arrays;
import java.util.Scanner;

public class Main {
    static int num;

    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        num = scan.nextInt()-1;
        int[]nums = new int[num];
        int[]nums2 = new int[num];
        int[]results = new int[num+1];
        for (int i = 0; i < num; i++) {
            nums [i] = scan.nextInt();
        }
        for (int i = 0; i < num; i++) {
            nums2[i] = scan.nextInt();
        }
        int i = 0;
        int res1 = 0;
        int res2 = 0;
        int counter = 0;

        while(true){
            if(i>=results.length-1){
                System.out.println("YES");
                System.out.println(Arrays.toString(results).replaceAll("[^0-9,\\.]", "")
                        .replaceAll("[^0-9\\.]"," ")
                        .trim());
                break;
            }else if(counter > 1000){
                System.out.println("NO");
                break;
            }
            if(i!=0) {
                int xor1 = results[i-1] | res2;
                int xor2 = res2 & results[i-1];
                if (xor1 == nums[i-1] && xor2 == nums2[i-1]) {
                    results[i] = res2;
                    //results[++i] = res2;
                    //res1 = 0;
                    res2 = 0;
                    i++;
                } else {
                    if(res2<3) {
                        res2++;
                        counter++;
                    }
                    else {
                        i = 0;
                        counter++;
                    }
                }
            }else if(i==0){
                int xor1 = res1|res2;
                int xor2 = res2&res1;
                if(xor1==nums[i]&&xor2==nums2[i]){
                    results[i] = res1;
                    results[++i] = res2;
                    res1 = 0;
                    res2 = 0;
                    i++;
                }else{
                    if(res2<3) {
                        res2++;
                        counter++;
                    }
                    else if(res1<3) {
                        res1++;
                        counter++;
                    }
                    else {
                        System.out.println("NO");
                        break;
                    }
            }
            }
        }
    }
}
