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.*;
public class AA{
    public static void main(String[] args) {
       MyScanner sk = new MyScanner();
       out = new PrintWriter(new BufferedOutputStream(System.out));
        int testcase = sk.nextInt();
        for (int i = 0; i < testcase; i++) {
            int n = sk.nextInt();
            int [][] a = new int [3][n];
            for (int j = 0; j < n; j++) {
                int input = sk.nextInt();
                a[0][j] = input;
            }
            for (int j = 0; j < n; j++) {
                int input = sk.nextInt();
                a[1][j] = input;
            }
            for (int j = 0; j < n; j++) {
                int input = sk.nextInt();
                a[2][j] = input;
            }
            int prev = a[0][0];
            int first = prev;
            int count = 0;
            out.print(prev + " ");
            for (int j = 1; j < n; j++) {
                if (j == n-1 && (a[count][j] == prev || a[count][j] == first)) {
                    do {
                        if (count == 2) count = 0;
                        else {
                            count++;
                        }
                    } while (a[count][j] == prev || a[count][j] == first);
                }
              while (a[count][j] == prev) {
                  if (count == 2) count = 0;
                  else {
                      count++;
                  }
              }
              prev = a[count][j];
              out.print(prev + " ");
            }
            out.println(); 
        }
    

       // Stop writing your solution here. -------------------------------------
        out.close();
    }

    //     int mins = s.length()+1212;
    //     for (Integer u = 1; u < a.length && a[u+1][0] != null; u++) {
    //         if (a[u-1][0].compareTo(a[u][0]) != 0 && a[u-1][0].compareTo(a[u+1][0]) != 0 && a[u][0].compareTo(a[u+1][0]) != 0) {
    //             mins = Math.min(mins, a[u][1] + 2);
    //         }
    //     }
    //     if (mins > s.length()) return 0;
    //     return mins;

    // }
    // public static int bsearch(int n, Vector<Integer> a, int k) {
    //     int l = 0, r = n + 1;
    //     while (l+1!= r) {
    //         int m = (l+r) / 2;
    //         if (a.get(m) <= k) {
    //             l = m;
    //         }else {
    //             r = m;
    //         }
    //     }
    //     return l;
    // }
    
    //-----------PrintWriter for faster output---------------------------------
    public static PrintWriter out;
       
    //-----------MyScanner class for faster input----------
    public static class MyScanner {
       BufferedReader br;
       StringTokenizer st;
  
       public MyScanner() {
          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;
       }
 
    }
    //--------------------------------------------------------
 }