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 tank {

    static final FastScanner fs = new FastScanner();

    static int[][][] cache = {
            {{0, 0}, {-1, -1}, {-1, -1}, {-1, -1}},
            {{1, 0}, {1, 1}, {-1, -1}, {-1, -1}},
            {{2, 0}, {-1, -1}, {2, 2}, {-1, -1}},
            {{2, 1}, {3, 1}, {3, 2}, {3, 3}}
    };

    static int[] t, a, b;

    public static void main(String[] args) {
        int n = fs.nextInt();
        a = fs.readArray(n - 1);
        b = fs.readArray(n - 1);
        t = new int[n];

        boolean yes = false;

        if(cache[a[0]][b[0]][0] == -1 || cache[a[0]][b[0]][1] == -1){
            System.out.println("NO");
            return;
        }

        t[0] = cache[a[0]][b[0]][0];
        if(fn(1, cache[a[0]][b[0]][1])) {
            yes = true;
        }

        if(!yes) {
            t[0] = cache[a[0]][b[0]][1];
            if(fn(1, cache[a[0]][b[0]][0]))
                yes = true;
        }

        if(!yes){
            if((a[0] == 3 || b[0] == 3) && (a[0] == 0 || b[0] == 0)){
                t[0] = 3;
                if(fn(1, 0)) {
                    yes = true;
                }

                if(!yes) {
                    t[0] = 0;
                    if(fn(1, 3))
                        yes = true;
                }
            }
        }

        if(yes){
            int ti = cache[a[n - 2]][b[n - 2]][0];
            int tip = cache[a[n - 2]][b[n - 2]][1];

            if(t[n - 2] == ti) {
                t[n - 1] = tip;
            }else{
                t[n - 1] = ti;
            }

            StringBuilder ans = new StringBuilder("YES\n");
            for(int i: t) ans.append(i).append(" ");

            System.out.println(ans);
        }else{
            System.out.println("NO");
        }
    }

    static boolean fn(int ind, int cur){
        if(ind == a.length) return true;

        int ti = cache[a[ind]][b[ind]][0];
        int tip = cache[a[ind]][b[ind]][1];

        if(ti == -1 || tip == -1){
            return false;
        }

        if(ti == cur || tip == cur){
            t[ind] = cur;

            if(ti == cur) cur = tip;
            else cur = ti;

            if(!fn(ind + 1, cur)){
                if((a[ind] == 3 || b[ind] == 3)
                        && (a[ind] == 0 || b[ind] == 0)){
                    ti = 3;
                    tip = 0;

                    if(ti == cur || tip == cur){
                        t[ind] = cur;

                        return fn(ind + 1, 3 - cur);
                    }
                }
                return false;
            }
            return true;
        }else{
            if((a[ind] == 3 || b[ind] == 3)
                    && (a[ind] == 0 || b[ind] == 0)){
                ti = 3;
                tip = 0;

                if(ti == cur || tip == cur){
                    t[ind] = cur;

                    return fn(ind + 1, 3 - cur);
                }
            }
            return false;
        }
    }

    static class FastScanner {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st=new StringTokenizer("");
        String next() {
            while (!st.hasMoreTokens())
                try {
                    st=new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            return st.nextToken();
        }

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

            return "";
        }

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

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

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

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

}

