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.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.util.ArrayList;
import java.util.Comparator;
import java.util.StringTokenizer;

public class A {

  public static void main(String[] args) {
    int T = 0;
    FastReader reader = new FastReader();
    BufferedWriter write = new BufferedWriter(new OutputStreamWriter(System.out));
    T = reader.nextInt();
    while (T-- > 0) {
      try {
        circleColoring(reader, write);
        write.flush();
      } catch (Exception ex) {
        System.out.println("invalid");
      }
    }
  }

  private static void circleColoring(FastReader reader, BufferedWriter writer) throws IOException {
    int ai = 0, bi = 0, ci = 0, n = 0, turn = 1, oi = 1;
    int[] a, b, c, out;
    n = reader.nextInt();
    a = new int[n];
    b = new int[n];
    c = new int[n];
    out = new int[n];
    StringBuilder output = new StringBuilder();
    for (int i = 0; i < n; i++) {
      a[i] = reader.nextInt();
    }
    for (int i = 0; i < n; i++) {
      b[i] = reader.nextInt();
    }
    for (int i = 0; i < n; i++) {
      c[i] = reader.nextInt();
    }
    out[0] = a[ai++];
    bi++;
    ci++;
    while (oi < n - 1) {
      switch (turn) {
        case 0:
          if (out[oi - 1] == a[ai]) {
            out[oi] = b[bi];
          } else {
            out[oi] = a[ai];
          }
          turn = 1;
          break;
        case 1:
          if (out[oi - 1] == b[bi]) {
            out[oi] = c[ci];
          } else {
            out[oi] = b[bi];
          }
          turn = 2;
          break;
        case 2:
          if (out[oi - 1] == c[ci]) {
            out[oi] = a[ai];
          } else {
            out[oi] = c[ci];
          }
          turn = 0;
          break;
      }
      ai++;
      bi++;
      ci++;
      oi++;
    }
    if (a[n - 1] != out[0] &&
      a[n - 1] != out[n - 2]) {
      out[oi] = a[n - 1];
    } else if (b[n - 1] != out[0] &&
      b[n - 1] != out[n - 2]) {
      out[oi] = b[n - 1];
    } else {
      out[oi] = c[n - 1];
    }
    for (int num : out) {
      output.append(num + " ");
    }
    output.deleteCharAt(output.length() - 1).append("\n");
    writer.write(output.toString());
  }

  static class FastReader
  {
    BufferedReader br;
    StringTokenizer st;

    public FastReader()
    {
      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;
    }
  }
}