Problem: 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.

Solution (in JAVA):

Solution:
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.FileReader;
import java.io.InputStreamReader;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        ACircleColoring solver = new ACircleColoring();
        solver.solve(1, in, out);
        out.close();
    }

    static class ACircleColoring {
        public void solve(int testNumber, Scanner sc, PrintWriter pw) {
            int t = 1;
            t = sc.nextInt();
            while (t-- > 0) {
                int n = sc.nextInt();
                int[] a = new int[n];
                int[] b = new int[n];
                int[] c = new int[n];
                for (int i = 0; i < n; i++) a[i] = sc.nextInt();
                for (int i = 0; i < n; i++) b[i] = sc.nextInt();
                for (int i = 0; i < n; i++) c[i] = sc.nextInt();
                int[] ans = new int[n];
                for (int i = 0; i < n; i++) {
                    if (i == 0) ans[i] = a[i];
                    else {
                        if (a[i] != ans[i - 1]) ans[i] = a[i];
                        else if (b[i] != ans[i - 1]) ans[i] = b[i];
                        else if (c[i] != ans[i - 1]) ans[i] = c[i];
                    }
                }
                if (ans[n - 1] == ans[0]) {
                    if (a[n - 1] != ans[0] && a[n - 1] != ans[n - 2]) ans[n - 1] = a[n - 1];
                    else if (b[n - 1] != ans[0] && b[n - 1] != ans[n - 2]) ans[n - 1] = b[n - 1];
                    else if (c[n - 1] != ans[0] && c[n - 1] != ans[n - 2]) ans[n - 1] = c[n - 1];
                }
                for (int i = 0; i < n; i++) pw.print(ans[i] + " ");
                pw.println();

            }
        }

    }

    static class Scanner {
        StringTokenizer st;
        BufferedReader br;

        public Scanner(FileReader r) {
            br = new BufferedReader(r);
        }

        public Scanner(InputStream s) {
            br = new BufferedReader(new InputStreamReader(s));
        }

        public String next() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return st.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

    }
}