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.
//package codeforces.practice.oneVone.Oct28_2020;

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

public class C {
    static InputReader in;
    static PrintWriter out;

    public static void main(String[] args) {
        //initReaderPrinter(true);
        initReaderPrinter(false);
        //solve(in.nextInt());
        solve(1);
    }

    static void solve(int testCnt) {
        for (int testNumber = 0; testNumber < testCnt; testNumber++) {
            int n = in.nextInt();
            Integer[] a = in.nextIntArray(n - 1);
            Integer[] b = in.nextIntArray(n - 1);
            int[][] aa = new int[n - 1][2], bb = new int[n - 1][2];
            for(int i = 0; i < n - 1; i++) {
                aa[i][0] = a[i] & 1;
                bb[i][0] = b[i] & 1;
                aa[i][1] = (a[i] >>> 1) & 1;
                bb[i][1] = (b[i] >>> 1) & 1;
            }
            int[] ans = null;
            boolean can = true;
            for(int start = 0; start <= 3; start++) {
                can = true;
                int[] t = new int[n];
                t[0] = start;
                for(int i = 1; i < n; i++) {
                    int[] prevD = new int[2], currD = new int[2];
                    boolean[] locked = new boolean[2];
                    prevD[0] = t[i - 1] & 1;
                    prevD[1] = t[i - 1] >>> 1;
                    for(int j = 0; j < 2; j++) {
                        //if a[i - 1]'s current digit is 0, then t[i - 1] and t[i]'s digits at the same position
                        //must be 0
                        if(aa[i - 1][j] == 0) {
                            if(prevD[j] != 0) {
                                can = false;
                                break;
                            }
                            currD[j] = 0;
                            locked[j] = true;
                        }
                        //else, if t[i - 1]'s digit is 0, then t[i]'s matching digit must be 1
                        else {
                            if(prevD[j] == 0) {
                                currD[j] = 1;
                                locked[j] = true;
                            }
                        }
                        //if b[i - 1]'s current digit is 0, then t[i - 1] and t[i] can not be 1 at the same time
                        if(bb[i - 1][j] == 0) {
                            if(prevD[j] != 0 && currD[j] != 0) {
                                can = false;
                                break;
                            }
                        }
                        //else, t[i - 1] and t[i] must be 1 at the same time
                        else {
                            if(prevD[j] == 0 || (currD[j] == 0 && locked[j])) {
                                can = false;
                                break;
                            }
                            currD[j] = 1;
                            locked[j] = true;
                        }
                    }
                    if(!can) {
                        continue;
                    }
                    t[i] = currD[0] + (currD[1] << 1);
                }
                if(can) {
                    ans = t;
                    break;
                }
            }
            if(can) {
                out.println("YES");
                for(int v : ans) {
                    out.print(v + " ");
                }
                out.println();
            }
            else {
                out.println("NO");
            }

        }
        out.close();
    }

    static void initReaderPrinter(boolean test) {
        if (test) {
            try {
                in = new InputReader(new FileInputStream("src/input.in"));
                out = new PrintWriter(new FileOutputStream("src/output.out"));
            } catch (IOException e) {
                e.printStackTrace();
            }
        } else {
            in = new InputReader(System.in);
            out = new PrintWriter(System.out);
        }
    }

    static class InputReader {
        BufferedReader br;
        StringTokenizer st;

        InputReader(InputStream stream) {
            try {
                br = new BufferedReader(new InputStreamReader(stream), 32768);
            } catch (Exception e) {
                e.printStackTrace();
            }
        }

        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());
        }

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

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

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }

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

        Long[] nextLongArray(int n) {
            Long[] a = new Long[n];
            for (int i = 0; i < n; i++) {
                a[i] = nextLong();
            }
            return a;
        }
    }
}


