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.
/******************************************************************************
 * Developed by Dmitriev Alexander on 09.10.18 18:56 for Technopolis          *
 * Last modified 09.10.18 18:56                                               *
 * https://www.vk.com/brainlux                                                *
 ******************************************************************************/

import java.math.BigDecimal;
import java.util.*;
import java.io.*;

public class Main {
    FastScanner in;
    PrintWriter out;


    public void solve() throws IOException {
        int n = in.nextInt();
        int[] a = new int[n];
        int[] b = new int[n];
        for (int i = 0; i < n - 1; i++)
            a[i] = in.nextInt();
        for (int i = 0; i < n - 1; i++)
            b[i] = in.nextInt();
        String s = "";
        if (b[0] == 3 && a[0] != 3) {
            No();
        }
        if (b[0] == 2 && a[0] == 1) {
            No();
        }
        if (b[0] == 1 && a[0] == 2) {
            No();
        }
        if (b[0] == 0)
            s += a[0] + " " + b[0] + " ";
        else if (a[0] == 3)
            s += b[0] + " " + a[0] + " ";
        else if (a[0] != b[0])
            No();
        else s += b[0] + " " + a[0] + " ";


        for (int i = 1; i < n - 1; i++) {
            if (b[i] == 3 && a[i] != 3) {
                No();
            }
            if (b[i] == 2 && a[i] == 1) {
                No();
            }
            if (b[i] == 1 && a[i] == 2) {
                No();
            }
            if (a[i] == b[i]) {
                if ((s.charAt(s.length() - 2) & a[i]) != b[i]) {
                    No();
                }
            }

            if (b[i] == 0) {
                if ((s.charAt(s.length() - 2) & a[i]) == 0) {
                    s += a[i];
                } else s += b[i];

            } else if (b[i] == 3)
                s += a[i] + " ";
            else s += b[i] + " ";
        }
        System.out.println("YES");
        System.out.println(s);

    }

    private void No() {
        System.out.println("NO");
        System.exit(0);
    }


    public void run() {
        try {
            in = new FastScanner(System.in);
            out = new PrintWriter(System.out);

            solve();

            out.close();
        } catch (IOException e) {
            e.printStackTrace();
        }
    }

    class FastScanner {
        BufferedReader br;
        StringTokenizer st;

        FastScanner(InputStream in) {
            br = new BufferedReader(new InputStreamReader(in));
        }

        String next() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        boolean ready() throws IOException {
            return br.ready();
        }
    }

    public static void main(String[] arg) {
        new Main().run();
    }
}