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.io.*;
import java.math.*;

public class Main 

{ 
	 static FastReader sc=new FastReader(); 
	 
		  public static void main(String[] args) 
    {
			  //StringBuffer sb=new StringBuffer("");
			    int t = i(); 
		        while (t-- > 0) 
				{ 	
		        	int n=i();
		        	int A[]=input(n);
		        	int B[]=input(n);
		        	int C[]=input(n);
		        	HashMap<Integer,Integer> map=new HashMap<>();
		        	int r[]=new int[n];
		        	r[0]=A[0];
		        	int prev=A[0];
		        	for(int i=1;i<n;i++) {
		        		if(prev!=A[i]) {
		        			prev=A[i];
		        			r[i]=A[i];
		        		}
		        		else if(prev!=B[i]) {
		        			prev=B[i];
		        			r[i]=B[i];
		        		}
		        		else if(prev!=C[i]) {
		        			prev=C[i];
		        			r[i]=C[i];
		        		}
		        	}
		        	for(int i : r) {
		        		System.out.print(i+" ");
		        	}
		        	System.out.println();
		        	
		        	
		        	
		        		
				}
		        //System.out.println(sb.toString());
    }
		  
		  
		  
		  
static int[] input(int n) {
	int A[]=new int[n];
	   for(int i=0;i<n;i++) {
		   A[i]=sc.nextInt();
	   }
	   return A;
   }
static long[] inputL(int n) {
	long A[]=new long[n];
	   for(int i=0;i<n;i++) {
		   A[i]=sc.nextLong();
	   }
	   return A;
   }
static String[] inputS(int n) {
	String A[]=new String[n];
	   for(int i=0;i<n;i++) {
		   A[i]=sc.next();
	   }
	   return A;
   }


static void input(int A[],int B[]) {
	   for(int i=0;i<A.length;i++) {
		   A[i]=sc.nextInt();
		   B[i]=sc.nextInt();
	   }
}
static long mod(long x) {
	 int mod=1000000007;
	  return ((x%mod + mod)%mod);
}

     static int i() {
    	 return sc.nextInt();
     }
     static String s() {
    	 return sc.next();
     }
     static long l() {
    	 return sc.nextLong();
     }  
     static void sort(int[] arr){
        int n = arr.length;
        Random rnd = new Random();
        for(int i=0; i<n; ++i){
            int tmp = arr[i];
            int randomPos = i + rnd.nextInt(n-i);
            arr[i] = arr[randomPos];
            arr[randomPos] = tmp;
        }
        Arrays.sort(arr);
     }
     static boolean prime(int n) 
	    { 
	        if (n <= 1) 
	            return false; 
	        if (n <= 3) 
	            return true; 
	        if (n % 2 == 0 || n % 3 == 0) 
	            return false; 
	        double sq=Math.sqrt(n);
	  
	        for (int i = 5; i <= sq; i = i + 6) 
	            if (n % i == 0 || n % (i + 2) == 0) 
	                return false; 
	        return true; 
	    } 
     static int gcd(int a, int b) 
     { 
         if (a == 0) 
             return b; 
         return gcd(b % a, a); 
     } 
  
        
        
    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; 
        } 
    } 
} 



