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 CuriosityHasNoLimits {
    static int[][] oTable=new int[4][4];
    static int[][] aTable=new int[4][4];
    public static void main(String[] args) throws IOException{
        Scanner in=new Scanner(System.in);
        int n=in.nextInt();

        int[] t=new int[n];
        int[] a=new int[n-1];
        int[] b=new int[n-1];
        for(int i=0;i<4;i++)
            for(int j=0;j<4;j++){
                oTable[i][j]=i|j;
                aTable[i][j]=i&j;
            }
        for(int i=0;i<n-1;i++)
            a[i]=in.nextInt();
        for(int i=0;i<n-1;i++)
            b[i]=in.nextInt();
        //find initial t's
        int i=0,j=0;
        for(;i<4;i++){
            j=find(i,a[0],b[0]);
            if(j!=-1)
                break;
        }
        if(j==-1){
            System.out.println("No");
            return;
        }

        t[0]=i;t[1]=j;
        String S=findT(t,n,a,b);
        if(S.equals("No")){
            t[0]=j;t[1]=i;
            S=findT(t,n,a,b);
        }
        if(S.equals("No") && a[0]==3 && b[0]==0){
            t[0]=1;t[1]=2;
            S=findT(t,n,a,b);
            if(S.equals("No")){
                t[0]=2;t[1]=1;
                S=findT(t,n,a,b);
            }
        }
        if(S.equals("No"))
            System.out.println("No");
        else{
            System.out.println("Yes");

            for(int k=0;k<n;k++)
                System.out.print(t[k]+" ");
        }


    }
    static String findT(int[] t,int n,int[] a,int[] b){
        String S="Yes";

        for(int i=2;i<n;i++){

            int idx=find(t[i-1],a[i-1],b[i-1]);
            if(idx==-1){
                S="No";
                break;
            }
            t[i]=idx;
        }
        return S;
    }


    static int find(int i,int or,int and){
        for(int j=0;j<4;j++)
            if(oTable[i][j]==or && aTable[i][j]==and)
                return j;
        return -1;
    }

}