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
{
    static int n;
    static int[] arr, brr, crr;
    static char[] s;
    public static void main(String[] args) throws IOException
    {
        Flash f = new Flash();
        int T = f.ni();
        for(int tc = 1; tc <= T; tc++){
            n = f.ni(); arr = f.arr(n); brr = f.arr(n); crr = f.arr(n);
            fn();
        }
    }
    
    static void fn()
    {
        int[] ans = new int[n];
        ans[0] = arr[0];
        for(int i = 1; i < n-1; i++){
            if(arr[i] != ans[i-1]) ans[i] = arr[i];
            else if(brr[i] != ans[i-1]) ans[i] = brr[i];
            else if(crr[i] != ans[i-1]) ans[i] = crr[i];
        }
        
        if(arr[n-1] != ans[n-2] && arr[n-1] != ans[0]) ans[n-1] = arr[n-1];
        else if(brr[n-1] != ans[n-2] && brr[n-1] != ans[0]) ans[n-1] = brr[n-1];
        else ans[n-1] = crr[n-1];
        print(ans);
    }
    
    static void sort(int[] a){
        List<Integer> A = new ArrayList<>();
        for(int i : a) A.add(i);
        Collections.sort(A);
        for(int i = 0; i < A.size(); i++) a[i] = A.get(i);
    }
    
    static int swap(int itself, int dummy){
        return itself;
    }
    
    static void print(int[] a){
        StringBuilder sb = new StringBuilder();
        for(int i = 0; i < a.length; i++) sb.append(a[i] + " ");
        System.out.println(sb);
    }
    
    static class Flash
    {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer("");
        
        String next(){
            while(!st.hasMoreTokens()){
                try{
                    st = new StringTokenizer(br.readLine());
                }catch(IOException e){
                    e.printStackTrace();
                }
            }
            
            return st.nextToken();
        }
        
        String nextLine(){
            String s = new String();
            try{
                s = br.readLine().trim();
            }catch(IOException e){
                e.printStackTrace();
            }
            
            return s;
        }
        
        int ni(){
            return Integer.parseInt(next());
        }
        
        int[] arr(int n){
            int[] a = new int[n];
            for(int i = 0; i < n; i++) a[i] = ni();
            return a;
        }

        long nl(){
            return Long.parseLong(next());
        }
    }
}