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.io.*;
import java.util.*;


public class A 
{	
	public static void main(String[] args)throws IOException 
	{	
		
		FastReader f=new FastReader();
		StringBuffer sb = new StringBuffer();
		
		int test=f.nextInt();
		while(test-->0)
		{
			int n=f.nextInt();
			int[] a=new int[n];
			for(int i=0;i<n;i++)
				a[i]=f.nextInt();
			
			int[] b=new int[n];
			for(int i=0;i<n;i++)
				b[i]=f.nextInt();
			
			int[] c=new int[n];
			for(int i=0;i<n;i++)
				c[i]=f.nextInt();
			
			int[] ans=new int[n];
			ans[0]=a[0];
			
			for (int i=1;i<n;i++) 
			{
				int prev=ans[i-1];
				if (i!=n-1) 
				{
					if(a[i]!=prev) 
						ans[i]=a[i];
					
					else if(b[i]!=prev) 
						ans[i]=b[i];
				}
				else 
				{
					int next=ans[0];
					if(a[i]!=prev && a[i]!=next) 
						ans[i]=a[i];
					
					else if(b[i]!=prev && b[i]!=next) 
						ans[i]=b[i];
					
					else if(c[i]!=prev && c[i]!=next) 
						ans[i]=c[i];
				}
			}
			for (int i:ans) 
				sb.append(i+" ");
			sb.append("\n");
		}
		System.out.println(sb);
	}
	
	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; 
	    } 
	} 
}
