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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.stream.IntStream;
import java.util.Arrays;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.util.LinkedList;
import java.io.InputStreamReader;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author dmytro.prytula
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        BLyubopitstvuNetPredela solver = new BLyubopitstvuNetPredela();
        solver.solve(1, in, out);
        out.close();
    }

    static class BLyubopitstvuNetPredela {
        int n;
        int[] a;
        private int[] b;
        int[][][] dp;
        boolean noAns = true;
        LinkedList<Integer> ll = new LinkedList<>();

        public void solve(int testNumber, InputReader in, PrintWriter out) {

            n = in.nextInt();
            a = in.nextArr(n - 1);
            b = in.nextArr(n - 1);

            dp = new int[n][4][4];


            for (int j = 0; j < 4; j++) {
                for (int k = 0; k < 4; k++) {
                    dp[0][j][k] = 1;
                }
            }

            for (int i = 1; i < n; i++) {
                for (int j = 0; j < 4; j++) {
                    for (int k = 0; k < 4; k++) {
                        if (a[i - 1] == (k | j) && b[i - 1] == (k & j)) {
                            dp[i][j][k] = 1;
                        }
                    }
                }
            }

//        traceMatrix(dp);

            for (int from = 0; from < 4; from++) {
                ll.addLast(from);
                dfs(1, from, out);
                if (!noAns) break;
                ll.removeLast();
            }

            if (noAns) {
                out.print("NO");
            }


        }

        private void dfs(int index, int from, PrintWriter out) {
            //System.out.println("dfs from" + from + " - cur " +  ll);


            if (index == n) {
                out.println("YES");
                for (Integer integer : ll) {
                    out.print(integer + " ");
                }
                out.println();
                noAns = false;
                return;
            }
            for (int to = 0; to < 4; to++) {
                if (dp[index][from][to] == 1) {
                    ll.addLast(to);
                    //      System.out.println("Exist path from " + from + " to "  + to);
                    dfs(index + 1, to, out);
                    //    System.out.println("End of the path " + from + " to "  + to);
                    ll.removeLast();
                }
            }
        }

    }

    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

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

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

        public int[] nextArr(int size) {
            return Arrays.stream(new int[size]).map(c -> nextInt()).toArray();
        }

    }
}

