Problem: 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.

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;
import java.util.StringTokenizer;

import java.util.*;

public class B {
	
	
	public static void main(String[]args) {
		FastReader fs=new FastReader();
		int t=fs.nextInt();
		for(int tt=0;tt<t;tt++) {
		  
		   int n=fs.nextInt(); 
			int a[]=new int[n];
			int b[]=new int[n];
			int c[]=new int[n];
			for(int i=0;i<n;i++) 
				a[i]=fs.nextInt();
			for(int i=0;i<n;i++)
				b[i]=fs.nextInt();
			for(int i=0;i<n;i++)	
				c[i]=fs.nextInt();
			
			int res[]=new int[n];
			int curr=-1;int fst=a[0];
			int i=0;
			while(i<n-1) {
				
				if(a[i]!=curr) {
					curr=a[i];
				res[i]=a[i];
					i++;
					
				
				}
				else if(b[i]!=curr) {
					curr=b[i];
					res[i]=b[i];
					
					i++;
				}
				else {
					curr=c[i];
					res[i]=c[i];
					i++;
				
				}	
			}
			if(a[n-1]!=curr&&a[n-1]!=fst)res[n-1]=a[n-1];
			else if(b[n-1]!=curr&&b[n-1]!=fst)res[n-1]=b[n-1];
			else res[n-1]=c[n-1];
			for(int ele:res)
				System.out.print(ele+" ");
		   
			System.out.println();
		}
		
	}
	static void solve(long a[],int n) {
		Map<Long,Long>hm=new HashMap<>();
		for(int i=0;i<n;i++) {
			long x=highestonebit(a[i]);
			if(!hm.containsKey(x))
				hm.put(x,Long.valueOf(1));
			else
				hm.put(x,hm.get(x)+1);
		}
		long result=0;
		for(long x:hm.values()) {
			if(x>=2)
				result+=x*(x-1)/2;
		}
		System.out.println(result);
		
		
	}
	static long highestonebit(long n) {
		if(n==0)return 0;
		else {
			long res=0;
			while(n>0) {
				n/=2;
				res++;
			}
			return res;
		}
	}
	
	
	static int solve15(long a[],int n) {
		Set<Long> hs=new HashSet<>();
		hs.add(Long.valueOf(0));
		long sum=0;int count=0;
		for(int i=0;i<n;i++) {
			sum+=a[i];
			if(hs.contains(sum)) {
				count++;
				hs.clear();
				hs.add(Long.valueOf(0));
				sum=a[i];
			}
			hs.add(sum);
		}
		
		
			
			
			
		
		return count;
	}
	static void solve11(int[]a,int n,int k) {
		int b[]=new int[n];
				Arrays.fill(b,-1);
		for(int i=0;i<n;i++) {
			if(a[i]*2<k)
				b[i]=0;
			else if(a[i]*2>k)
				b[i]=1;
			
			}
		int z=0;
		for(int i=0;i<n;i++) {
			if(b[i]==-1) {
				b[i]=z;
				z^=1;
			
		}}
	
			
		for(int ele:b)
			System.out.print(ele+" ");
		
				
			}
		
	
	
	public static void solve1(String s) {
		String []ss=s.split("0");
		List<String>l=new ArrayList<>();
		for(String st:ss) {
			if(st!=null||st.length()!=0)
				l.add(st);
		}
		int arr[]=new int[l.size()];
		
		for(int i=0;i<l.size();i++) 
			arr[i]=l.get(i).length();
			
		Arrays.sort(arr);	
		
		int count=0;
		for(int i=arr.length-1;i>=0;i-=2)
			count+=arr[i];
			
				
			
			System.out.println(count);
		
	
	
	}
	

	
	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; 
		} 
	} 


}