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 Task {

    public static void main(String[] args) throws IOException {

        new Task().go();
    }

    PrintWriter out;
    Reader in;
    BufferedReader br;

    Task() throws IOException {

        try {

            //br = new BufferedReader( new FileReader("input.txt") );
            in = new Reader("input.txt");
            out = new PrintWriter( new BufferedWriter(new FileWriter("output.txt")) );
        }
        catch (Exception e) {

            //br = new BufferedReader( new InputStreamReader( System.in ) );
            in = new Reader();
            out = new PrintWriter( new BufferedWriter(new OutputStreamWriter(System.out)) );
        }
    }

    void go() throws IOException {

        int t = 1;
        while (t > 0) {
            solve();
            //out.println();
            t--;
        }

        out.flush();
        out.close();
    }

    int inf = 2000000000;
    int mod = 1000000007;
    double eps = 0.000000001;

    int n;
    int m;

    int[] a;
    ArrayList<Integer>[] g;
    void solve() throws IOException {

        int n = in.nextInt();
        int[] a = in.nextIntArray(n - 1);
        int[] b = in.nextIntArray(n - 1);

        int[][] dp = new int[4][n];
        boolean[][] exist = new boolean[4][n];
        for (int i = 0; i < 4; i++) {
            exist[i][0] = true;
            Arrays.fill(dp[i], -1);
        }
        for (int i = 1; i < n; i++) {
            for (int j = 0; j < 4; j++)
                if (exist[j][i - 1])
                    for (int k = 0; k < 4; k++)
                        if ((k | j) == a[i - 1] && (k & j) == b[i - 1]) {
                            //System.err.println(j + " " + k);
                            exist[k][i] = true;
                            dp[k][i] = j;
                        }
        }

        int cur = -1;
        for (int i = 0; i < 4; i++)
            if (exist[i][n - 1])
                cur = i;

        if (cur == -1) {
            out.print("NO");
            return;
        }

        out.println("YES");
        ArrayList<Integer> ans = new ArrayList<>();
        for (int i = n - 1; i >= 0; i--) {
            ans.add(cur);
            cur = dp[cur][i];
        }
        for (int i = 0; i < n; i++)
            out.print(ans.get(n - i - 1) + " ");
    }

    int gcd(int a, int b) {
        if (b == 0)
            return a;
        return gcd(b, a % b);
    }

    class Pair implements Comparable<Pair>{
        int a;
        int b;

        Pair(int a, int b) {
            this.a = a;
            this.b = b;
        }

        public int compareTo(Pair p) {
            if (a > p.a) return 1;
            if (a < p.a) return -1;
            if (b > p.b) return 1;
            if (b < p.b) return -1;
            return 0;
        }
    }

    class Item {

        int a;
        int b;
        int c;

        Item(int a, int b, int c) {
            this.a = a;
            this.b = b;
            this.c = c;
        }

    }

    class Reader {

        BufferedReader  br;
        StringTokenizer tok;

        Reader(String file) throws IOException {
            br = new BufferedReader( new FileReader(file) );
        }

        Reader() throws IOException {
            br = new BufferedReader( new InputStreamReader(System.in) );
        }

        String next() throws IOException {

            while (tok == null || !tok.hasMoreElements())
                tok = new StringTokenizer(br.readLine());
            return tok.nextToken();
        }

        int nextInt() throws NumberFormatException, IOException {
            return Integer.valueOf(next());
        }

        long nextLong() throws NumberFormatException, IOException {
            return Long.valueOf(next());
        }

        double nextDouble() throws NumberFormatException, IOException {
            return Double.valueOf(next());
        }


        String nextLine() throws IOException {
            return br.readLine();
        }

        int[] nextIntArray(int n) throws IOException {
            int[] a = new int[n];
            for (int i = 0; i < n; i++)
                a[i] = nextInt();
            return a;
        }

        ArrayList<Integer>[] nextGraph(int n, int m) throws IOException {
            ArrayList<Integer>[] g = new ArrayList[n];
            for (int i = 0; i < n; i++)
                g[i] = new ArrayList<>();
            for (int i = 0; i < m; i++) {
                int x = nextInt() - 1;
                int y = nextInt() - 1;
                g[x].add(y);
                g[y].add(x);
            }

            return g;
        }

    }

}