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.util.Scanner;

/**
 * Created by User on 21.10.2018.
 */
public class Task2 {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        String[] a = new String[n-1];
        String[] b = new String[n-1];
        String[] t = new String[n];
        for (int i=0; i<n; i++){
            t[i]="nn";
        }
        for (int i=0; i<n-1; i++){
            int e=sc.nextInt();
            switch(e){
                case 0: a[i]="00";
                break;
                case 1: a[i]="01";
                break;
                case 2: a[i]="10";
                break;
                case 3: a[i]="11";
                break;
            }
        }
        for (int i=0; i<n-1; i++){
            int e=sc.nextInt();
            switch(e){
                case 0: b[i]="00";
                break;
                case 1: b[i]="01";
                break;
                case 2: b[i]="10";
                break;
                case 3: b[i]="11";
                break;
            }
        }
        for (int i=0; i<n-1; i++){
            char e1 = a[i].charAt(0);
            char e2 = b[i].charAt(0);
            char f1 = a[i].charAt(1);
            char f2 = b[i].charAt(1);
            if (e1=='0'){
                if (t[i].charAt(0)!='1'){
                    t[i]="0"+t[i].charAt(1);
                    t[i+1]="0"+t[i+1].charAt(1);
                }
                else{
                    System.out.println("NO");
                    System.exit(0);
                }
            }
            if (e2=='1'){
                if (t[i].charAt(0)!='0'){
                    t[i]="1"+t[i].charAt(1);
                    t[i+1]="1"+t[i+1].charAt(1);
                }
                else{
                    System.out.println("NO");
                    System.exit(0);
                }
            }
            if (f1=='0'){
                if (t[i].charAt(1)!='1'){
                    t[i]=t[i].charAt(0)+"0";
                    t[i+1]=t[i+1].charAt(0)+"0";
                }
                else{
                    System.out.println("NO");
                    System.exit(0);
                }
            }
            if (f2=='1'){
                if (t[i].charAt(1)!='0'){
                    t[i]=t[i].charAt(0)+"1";
                    t[i+1]=t[i+1].charAt(0)+"1";
                }
                else{
                    System.out.println("NO");
                    System.exit(0);
                }
            }
        }
        for (int i=0; i<n; i++){
            if(t[i].contains("n")) t[i]=t[i].replace('n', '0');
        }
        System.out.println("YES");
        for (int i=0; i<n; i++){
            switch(t[i]){
                case "00":
                    System.out.print("0 ");
                    break;
                case "01":
                    System.out.print("1 ");
                    break;
                case "10":
                    System.out.print("2 ");
                    break;
                case "11":
                    System.out.print("3 ");
                    break;
            }
        }
    }
}
