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 Main {
    String filename = "";

    FastScanner in;
    PrintWriter out;

    void solve() {
        int n = in.nextInt();
        int[] a = new int[n-1];
        int[] b = new int[n-1];
        int[] t = new int[n];
        int k = 0;
        for (int i = 0; i < n; i++) {
            t[i] = 0;
        }
        for (int i = 0; i < n-1; i++) {
            a[i] = in.nextInt();
        }
        for (int i = 0; i < n-1; i++) {
            b[i] = in.nextInt();
        }
        for (int i = 0; i < n-1; i++) {
            if(a[i]==b[i]&&(t[i]==0||t[i]==a[i])){
                t[i] = a[i];
                t[i+1] = b[i];
            }else if(b[i]==0){
                if(t[i]==b[i]){
                    t[i+1] = a[i];
                }else if(t[i]==a[i]){
                    t[i+1] = b[i];
                }else if(a[i]==3){
                    if(t[i]==2){
                        t[i+1] = 1;
                    }else if(t[i]==1){
                        t[i+1] = 2;
                    }else{
                        k++;
                    }
                }else{
                    k++;
                }
            }else if(b[i]==2&&a[i]==3||b[i]==1&&a[i]==3){
                if(t[i]==a[i]){
                    t[i+1] = b[i];
                }else if(t[i]==b[i]){
                    t[i+1] = a[i];
                }else if(t[i]==0){
                    if(i==0&&a[i+1]==3){
                        t[i+1] = a[i];
                        t[i] = b[i];
                    }else if(i==0){
                        t[i] = a[i];
                        t[i+1] = b[i];
                    }else{
                        k++;
                    }
                }else{
                    k++;
                }
            }else{
                k++;
            }
        }
        if(k>0){
            out.print("NO");
        }else{
            out.println("YES");
            for (int i = 0; i < n; i++) {
                out.print(t[i]+" ");
            }
        }
    }

    void run() throws IOException {
        InputStream input = System.in;
        OutputStream output = System.out;
        try {
            File f = new File(filename + ".in");
            if (f.exists() && f.canRead()) {
                input = new FileInputStream(f);
                output = new FileOutputStream(filename + ".out");
            }
        } catch (IOException e) {
        }
        in = new FastScanner(input);
        out = new PrintWriter(new BufferedOutputStream(output));
        solve();
        in.close();
        out.close();
    }

    public static void main(String[] args) throws IOException {
        Locale.setDefault(Locale.US);
        new Main().run();
    }

    class FastScanner implements Closeable {
        private BufferedReader br;
        private StringTokenizer tokenizer;

        public FastScanner(InputStream stream) throws FileNotFoundException {
            br = new BufferedReader(new InputStreamReader(stream));
        }

        public boolean hasNext() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    String s = br.readLine();
                    if (s == null) {
                        return false;
                    }
                    tokenizer = new StringTokenizer(s);
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.hasMoreTokens();
        }

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

        public String nextLine() {
            if (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    return br.readLine();
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken("\n");
        }

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

        public long nextLong() {
            return Long.parseLong(next());
        }

        public double nextDouble() {
            return Double.parseDouble(next());
        }

        @Override
        public void close() throws IOException {
            br.close();
        }
    }
}