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.
//package com.company;

import java.io.BufferedReader;
import java.io.*;
import java.io.IOException;
import java.io.PrintWriter;
import java.util.*;

public class solution {
    BufferedReader br;
    PrintWriter out;

    private void solve() {
        int t = ni();
        while(t-- > 0) {
            int n = ni();
            int[] ans = new int[n];
            HashSet<Integer>[] a = new HashSet[n];
            for(int i = 0; i < n; i++) a[i] = new HashSet<>();
            for(int j = 0; j < 3; j++) {
                for (int i = 0; i < n; i++) {
                    a[i].add(ni());
                }
            }
            for(int i: a[0]) {
                int prev = i;
                ans[0] = i;
                boolean found = find(a, prev, 1, ans);
                if(found) break;
            }
            for (int i = 0; i < n-1; i++) {
                out.print(ans[i] + " ");
            }
            out.println(ans[n-1]);
        }
    }

    private boolean find(HashSet<Integer>[] a, int prev, int i, int[] ans) {
        if(i >= a.length) return true;
        for(int val: a[i]) {
            if(val != prev) {
                if(i == a.length-1) {
                    if(val == ans[0]) continue;
                }
                ans[i] = val;
                if(find(a, val, i+1, ans)) {
                    return true;
                }
            }
        }
        return false;
    }

    // -------- I/O Template -------------

    char nc() {
        return ns().charAt(0);
    }

    String nLine() {
        try {
            return br.readLine();
        } catch(IOException e) {
            return "-1";
        }
    }

    double nd() {
        return Double.parseDouble(ns());
    }

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

    int ni() {
        return Integer.parseInt(ns());
    }

    StringTokenizer ip;

    String ns() {
        while(ip == null || !ip.hasMoreTokens()) {
            try {
                ip = new StringTokenizer(br.readLine());
            } catch(IOException e) {
                throw new InputMismatchException();
            }
        }
        return ip.nextToken();
    }

    void run() {
        br = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(System.out);
        solve();
        out.flush();
    }

    public static void main(String[] args) {
        new solution().run();
    }

}