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.*;
public class Main {
    public static int[] arr(int[] a, int[] b, int start) {
        int[] seq = new int[a.length + 1];
        seq[0] = start;
        for (int i = 0; i < a.length; i++) {
            if (seq[i] == 0) {
                if (a[i] == 0 && b[i] == 0) seq[i + 1] = 0;
                if (a[i] == 1 && b[i] == 0) seq[i + 1] = 1;
                if (a[i] == 0 && b[i] == 1) return null;
                if (a[i] == 1 && b[i] == 1) return null;
            } else {
                if (a[i] == 0 && b[i] == 0) return null;
                if (a[i] == 1 && b[i] == 0) seq[i + 1] = 0;
                if (a[i] == 0 && b[i] == 0) return null;
                if (a[i] == 1 && b[i] == 1) seq[i + 1] = 1;
            }
        }
        return seq;
    }
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int[][] a = new int[2][n - 1];
        int[][] b = new int[2][n - 1];
        for (int i = 0; i < n - 1; i++) {
            String s = Integer.toBinaryString(in.nextInt());
            if (s.length() == 1) s = "0" + s;
            a[0][i] = s.charAt(0) - '0';
            a[1][i] = s.charAt(1) - '0';
        }
        for (int i = 0; i < n - 1; i++) {
            String s = Integer.toBinaryString(in.nextInt());
            if (s.length() == 1) s = "0" + s;
            b[0][i] = s.charAt(0) - '0';
            b[1][i] = s.charAt(1) - '0';
        }
        int[] A = arr(a[0], b[0], 0);
        if (A == null) {
            A = arr(a[0], b[0], 1);
        }
        int[] B = arr(a[1], b[1], 0);
        if (B == null) {
            B = arr(a[1], b[1], 1);
        }
        if (A == null || B == null) {
            System.out.println("NO");
        } else {
            System.out.println("YES");
            for (int i = 0; i < n; i++) {
                System.out.print((A[i] * 2 + B[i]) + " ");
            }
            System.out.println();
        }
    }
}