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.io.*;
import java.util.*;

public class A {
    static int n, a[], b[], an = 0;
    static List<Integer>[] list;

    void run(FastIO io) {
        n = io.nextInt();
        a = new int[n - 1];
        b = new int[n - 1];
        list = new List[2];

        for (int i = 0; i < 2; i++) {
            list[i] = new ArrayList<>();
        }
        for (int i = 0; i < n - 1; i++) {
            a[i] = io.nextInt();
        }
        for (int i = 0; i < n - 1; i++) {
            b[i] = io.nextInt();
        } //получаем Аi и Bi

        boolean end = (a[0] & b[0]) < b[0]; //если уже первая пара невозможна

        if (n == 2 || end) //если всего 1 пара
            if (end)
                io.println("NO");
            else
                io.println("YES" + "\r\n" + a[0] + " " + b[0]);

        else { //если пар > 1

            for (int i = 1; i < n - 1; i++) {
                if ((a[i] & b[i]) < b[i] || a[i] != a[i - 1] && b[i] != a[i - 1] && a[i] != b[i - 1] && b[i] != b[i - 1]) //проверка пар на состыковку
                    end = true;
            }

            if (!end) { //если состыковка теоретически возможна
                list[0].add(a[0]);
                list[0].add(b[0]);
                list[1].add(b[0]);
                list[1].add(a[0]);

                if (cheker(1, 0)) an = 0;
                else if (cheker(1, 1)) an = 1; //строим два листа
                else end = true;
            }

            if (end)
                io.println("NO");
            else {
                io.println("YES");
                for (int i = 0; i < n; i++) {
                    io.print(list[an].get(i) + " ");
                }
            }
        }
    }

    public boolean cheker(int i, int m) {
        if (i == n - 1) return true;
        int last = list[m].get(i);
        if (last == a[i] || last == b[i])
            list[m].add(last != a[i] ? a[i] : b[i]);
        else
            return false;
        return cheker(i + 1, m);
    }

    public static void main(String[] args) throws InterruptedException {
        try (FastIO io = new FastIO(System.in, System.out)) {
            Thread t = new Thread(null, () -> new A().run(io), "", (long) 3e7);
            t.start();
            t.join();
        }
    }

    static class FastIO extends PrintWriter {
        BufferedReader br;
        StringTokenizer stok;

        FastIO(InputStream is, OutputStream os) {
            super(os, false);
            br = new BufferedReader(new InputStreamReader(is));
            stok = new StringTokenizer("");
        }

        String nextToken() {
            while (!stok.hasMoreTokens()) {
                String s;
                try {
                    s = br.readLine();
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
                if (s == null) {
                    return null;
                }
                stok = new StringTokenizer(s);
            }
            return stok.nextToken();
        }

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

        long nextLong() {
            return Long.parseLong(nextToken());
        }

        double nextDouble() {
            return Double.parseDouble(nextToken());
        }
    }
}