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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Scanner;
import java.util.StringTokenizer;

public class CodeForces {
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter w = new PrintWriter(System.out);
        int n = Integer.parseInt(br.readLine());
        int[] a = new int[n - 1];
        int[] b = new int[n - 1];
        String s = br.readLine();
        StringTokenizer st = new StringTokenizer(s);
        for (int i = 0; i < n - 1; ++i) {
            a[i] = Integer.parseInt(st.nextToken());
        }
        s = br.readLine();
        st = new StringTokenizer(s);
        for (int i = 0; i < n - 1; ++i) {
            b[i] = Integer.parseInt(st.nextToken());
        }
        int[] t = new int[n];
        boolean[] last = new boolean[n];
        for (int i = 0; i < n - 1; ++i) {
            if (i == 0) {
                if (a[i] == b[i]) {
                    t[i] = a[i];
                    t[i + 1] = a[i];
                    last[i] = true;
                } else if (a[i] == 0 || b[i] == 3) {
                    w.println("NO");
                    w.close();
                    return;
                } else {
                    if (a[i] == 2 || a[i] == 1) {
                        if (b[i] == 3 - a[i]) {
                            w.println("NO");
                            w.close();
                            return;
                        } else {
                            t[i] = a[i];
                            t[i + 1] = 0;
                        }
                    } else {
                        t[i] = 3;
                        t[i + 1] = b[i];
                    }
                }
            } else {
                if (a[i] == b[i]) {
                    if (t[i] != a[i] && (last[i - 1] || t[i - 1] != a[i])) {
                        w.println("NO");
                        w.close();
                        return;
                    } else if (t[i] != a[i]) {
                        int per = t[i - 1];
                        t[i - 1] = t[i];
                        t[i] = per;
                    }
                    t[i + 1] = a[i];
                    last[i] = true;
                } else if (a[i] == 0 || b[i] == 3) {
                    w.println("NO");
                    w.close();
                    return;
                } else {
                    if (a[i] == 2 || a[i] == 1) {
                        if (b[i] == 3 - a[i]) {
                            w.println("NO");
                            w.close();
                            return;
                        } else {
                            if ((t[i] != a[i] && t[i] != 0) && (last[i - 1] || (t[i - 1] != a[i] && t[i - 1] != 0))) {
                                w.println("NO");
                                w.close();
                                return;
                            } else if (t[i] != a[i] && t[i] != 0) {
                                int per = t[i - 1];
                                t[i - 1] = t[i];
                                t[i] = per;
                                last[i] = true;
                            }
                            t[i + 1] = a[i] - t[i];
                        }
                    } else {
                        if ((t[i] != b[i] && t[i] != a[i]) && (last[i - 1] || (t[i - 1] != b[i] && t[i - 1] != 3))) {
                            w.println("NO");
                            w.close();
                            return;
                        } else if (t[i] != b[i] && t[i] != a[i]) {
                            int per = t[i - 1];
                            t[i - 1] = t[i];
                            t[i] = per;
                            last[i] = true;
                        }
                        t[i + 1] = (b[i] + a[i]) - t[i];
                    }
                }
            }
        }
        w.println("YES");
        for (int i = 0; i < n; ++i) {
            w.print(t[i] + " ");
        }
        w.close();
    }
}
