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.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Set;
import java.util.StringTokenizer;
import java.util.TreeMap;
import java.util.TreeSet;

public class Solution {
	private BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
	private BufferedWriter out = new BufferedWriter(new OutputStreamWriter(System.out));
	private StringTokenizer st;
	private Long MOD = 1000000007L;

	public static void main(String[] args) throws Exception {
		new Solution().solve();
	}

	private void solve() throws Exception {

		int testCases = 1;
		testCases = Integer.parseInt(in.readLine().trim());
		for (int i = 1; i <= testCases; i++) {
			solve(i);
		}

		out.flush();
		out.close();
	}

	private void solve(int testCaseNumber) throws Exception {

		int n = ri();

		int[] a = ria(n);
		int[] b = ria(n);
		int[] c = ria(n);

		if (n == 1) {
			out.write(a[0] + "\n");
			return;
		}
		int[] ans = new int[n];
		int pre = a[n - 1];
		for (int i = 0; i < n; i++) {

			ans[i] = a[i];
			if (ans[i] == pre)
				ans[i] = b[i];
			pre = ans[i];
		}
		if (ans[n - 1] == ans[0] || ans[n - 1] == ans[n - 2])
			ans[n - 1] = a[n - 1];
		if (ans[n - 1] == ans[0] || ans[n - 1] == ans[n - 2])
			ans[n - 1] = b[n - 1];
		if (ans[n - 1] == ans[0] || ans[n - 1] == ans[n - 2])
			ans[n - 1] = c[n - 1];

		for (int x : ans)
			out.write(x + " ");

		out.write("\n");

	}

	class Node implements Comparable<Node> {
		int s;
		int e;

		public Node(int i, int t) {
			// TODO Auto-generated constructor stub
			s = i;
			e = t;
		}

		@Override
		public int compareTo(Node o) {
			return s - o.s;
		}

	}

	// utilities below

	// IO UTILS
	String line() throws IOException {
		return in.readLine();
	}

	String l() throws IOException {
		return in.readLine();
	}

	StringTokenizer tokens() throws IOException {
		return new StringTokenizer(line());
	}

	long l(String s) {
		return Long.parseLong(s);
	}

	long rl() throws IOException {
		return l(line());
	}

	long[] rla(int n) throws IOException {
		long[] arr = new long[n];
		st = tokens();

		for (int i = 0; i < n; i++)
			arr[i] = l(st.nextToken());

		return arr;
	}

	int i(String s) {
		return Integer.parseInt(s);
	}

	int ri() throws IOException {
		return i(line());
	}

	int[] ria(int n) throws IOException {
		int[] arr = new int[n];
		st = tokens();

		for (int i = 0; i < n; i++)
			arr[i] = i(st.nextToken());

		return arr;
	}

	String rs() throws IOException {
		return line();
	}

	String[] rsa(int n) throws IOException {
		String s[] = new String[n];

		st = tokens();

		for (int i = 0; i < n; i++)
			s[i] = st.nextToken();

		return s;
	}
	// IO UTILS END

	// MATH UTILS

	long gcd(long a, long b) {
		return b == 0 ? a : gcd(b, a % b);
	}
	// MATH UTILS END
}