You are given three sequences: a_1, a_2, …, a_n; b_1, b_2, …, b_n; c_1, c_2, …, c_n.

For each i, a_i ≠ b_i, a_i ≠ c_i, b_i ≠ c_i.

Find a sequence p_1, p_2, …, p_n, that satisfy the following conditions:

  * p_i ∈ \\{a_i, b_i, c_i\}
  * p_i ≠ p_{(i mod n) + 1}.



In other words, for each element, you need to choose one of the three possible values, such that no two adjacent elements (where we consider elements i,i+1 adjacent for i<n and also elements 1 and n) will have equal value.

It can be proved that in the given constraints solution always exists. You don't need to minimize/maximize anything, you need to find any proper sequence.

Input

The first line of input contains one integer t (1 ≤ t ≤ 100): the number of test cases.

The first line of each test case contains one integer n (3 ≤ n ≤ 100): the number of elements in the given sequences.

The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 100).

The third line contains n integers b_1, b_2, …, b_n (1 ≤ b_i ≤ 100).

The fourth line contains n integers c_1, c_2, …, c_n (1 ≤ c_i ≤ 100).

It is guaranteed that a_i ≠ b_i, a_i ≠ c_i, b_i ≠ c_i for all i.

Output

For each test case, print n integers: p_1, p_2, …, p_n (p_i ∈ \\{a_i, b_i, c_i\}, p_i ≠ p_{i mod n + 1}).

If there are several solutions, you can print any.

Example

Input


5
3
1 1 1
2 2 2
3 3 3
4
1 2 1 2
2 1 2 1
3 4 3 4
7
1 3 3 1 1 1 1
2 4 4 3 2 2 4
4 2 2 2 4 4 2
3
1 2 1
2 3 3
3 1 2
10
1 1 1 2 2 2 3 3 3 1
2 2 2 3 3 3 1 1 1 2
3 3 3 1 1 1 2 2 2 3


Output


1 2 3
1 2 1 2
1 3 4 3 2 4 2
1 3 2
1 2 3 1 2 3 1 2 3 2

Note

In the first test case p = [1, 2, 3].

It is a correct answer, because:

  * p_1 = 1 = a_1, p_2 = 2 = b_2, p_3 = 3 = c_3 
  * p_1 ≠ p_2 , p_2 ≠ p_3 , p_3 ≠ p_1 



All possible correct answers to this test case are: [1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 1, 2], [3, 2, 1].

In the second test case p = [1, 2, 1, 2].

In this sequence p_1 = a_1, p_2 = a_2, p_3 = a_3, p_4 = a_4. Also we can see, that no two adjacent elements of the sequence are equal.

In the third test case p = [1, 3, 4, 3, 2, 4, 2].

In this sequence p_1 = a_1, p_2 = a_2, p_3 = b_3, p_4 = b_4, p_5 = b_5, p_6 = c_6, p_7 = c_7. Also we can see, that no two adjacent elements of the sequence are equal.
import java.util.*;
import java.util.Scanner;
import java.io.*;
import javax.lang.model.util.ElementScanner6; 
import static java.lang.System.out;


public class A1408T
{

    public static void main(String args[])
    {  

        FastReader in=new FastReader();
        PrintWriter pr = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
        int tc=1;
     
        tc=in.nextInt();
        while(tc-->0)
        {
            
           int n=in.nextInt();

           int arr[]=new int[n];

           for(int i=0;i<n;i++)arr[i]=in.nextInt();

           int brr[]=new int[n];

           for(int i=0;i<n;i++)brr[i]=in.nextInt();

           int crr[]=new int[n];

           for(int i=0;i<n;i++)crr[i]=in.nextInt();

           pr.print(arr[0]+" ");
           int last=arr[0];
           for(int i=1;i<n-1;i++)
           {
               if(arr[i]!=last)
               {
                   last=arr[i];
                   pr.print(arr[i]+" ");
               }
               else if(brr[i]!=last)
               {
                   last=brr[i];
                   pr.print(brr[i]+" ");
               }
               else
               {
                   last=crr[i];
                   pr.print(crr[i]+" ");

               }
            }
            if(arr[n-1]!=arr[0]&&arr[n-1]!=last)
               {
          
                   pr.print(arr[n-1]+" ");
               }
               else if(brr[n-1]!=arr[0]&&brr[n-1]!=last)
               {
            
                   pr.print(brr[n-1]+" ");
               }
               else
               {
                   pr.print(crr[n-1]+" ");

               }

               pr.println();
          


        }
        pr.flush();
    }

    static class Pair implements Comparable<Pair> {
        int val, idx;
 
        public Pair(int val, int idx) {
            this.val = val;
            this.idx = idx + 1;
        }
 
        @Override
        public int compareTo(Pair o) {
            return val - o.val;
        }
    }
    
    static int gcd(int a,int b)
    {
        
        if(a==0)
            return b;
        return gcd(b%a,a);
    }

    static int lcm(int a,int b)
    {
        return (a*b)/gcd(a,b);
    }


    static boolean prime[];

    static void sieveofe()
    {
        int n=1000000;
      
        prime=new boolean[n+1];
        Arrays.fill(prime,true);
        prime[1]=false;
        for(int i=2;i*i<=n;i++)
        {
            if(prime[i]==true)
            {
                for(int j=i*i;j<=n;j+=i)
                {
                    prime[j]=false;
                }
            }
        }
    }

    
    //to always sort via merge sort, as sorting a primitive ds uses quick sort 
    //which can have O(n^2) complexity, thus not optimal
    static void sort(long[] a) {
		ArrayList<Long> l = new ArrayList<>();
		for (long i : a)
			l.add(i);
		Collections.sort(l);
		for (int i = 0; i < a.length; i++)
			a[i] = l.get(i);
	}
    
    
	static void sort(int[] a) {
		ArrayList<Integer> l = new ArrayList<>();
		for (int i : a)
			l.add(i);
		Collections.sort(l);
		for (int i = 0; i < a.length; i++)
			a[i] = l.get(i);
    }

    
    static class FastReader
    { 
        BufferedReader br; 
        StringTokenizer st; 

        public FastReader() 
        { 
            br = new BufferedReader(new InputStreamReader(System.in)); 
        } 

        String next() 
        { 
            while (st == null || !st.hasMoreElements()) 
            { 
                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; 
        } 
    } 
}