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.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;

public class A {

	
	static Boolean[][] dp = new Boolean[4][100009];
	static Integer[][] chose = new Integer[4][100009];
	
	static int[] a;
	static int[] b;
	

	static int n;
	
	public static void main(String[] args) throws IOException {
		
        PrintWriter out = new PrintWriter(System.out);
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        
        n = Integer.parseInt(in.readLine());

        a = new int[n];
        b = new int[n];
        
        String[] line = in.readLine().split("\\s+");
        for(int i=0; i<n-1; i++) {
        	a[i] = Integer.parseInt(line[i]);
        }
        
        line = in.readLine().split("\\s+");
        for(int i=0; i<n-1; i++) {
        	b[i] = Integer.parseInt(line[i]);
        }
        
        for(int i=0; i<=3; i++) {
        	if(rec(i, 0)) {
        		out.println("YES");
        		
        		out.print(i + " ");
        		
        		int prev = i;
        		for(int j=0; j<n-1; j++) {
        			prev = chose[prev][j];
        			
        			out.print(prev + " ");
        		}
        		
        		out.close();
        		return;
        	}
        }
        
        out.println("NO");
        
        out.close();
	
	}

	private static boolean rec(int prev, int i) {
		if(i == n) {
			return true;
		}
		else if(dp[prev][i] != null) {
			return dp[prev][i];
		}
		else {
			
			
			for(int j=0; j<=3; j++) {
				if( (j | prev) == a[i] && (j & prev) == b[i]) {
					if(rec(j, i+1)){ 
						dp[prev][i] = true;
						chose[prev][i] = j;
						
						return true;
					}
				}
			}
			
			dp[prev][i] = false;
			return false;

			
			
		}
	}

}
