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.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.HashMap;
import java.util.Random;
import java.util.StringTokenizer;
import java.util.*;
import static java.lang.Double.parseDouble;
import static java.lang.Integer.parseInt;
import static java.lang.Long.parseLong;
import static java.lang.Math.*;
import static java.lang.System.in;
import static java.lang.System.out;

public class Main {
    static FastScanner in = new FastScanner();
    static PrintWriter out = new PrintWriter(System.out);

    public static void main(String[] args) {
        int test = ri();
        while(test-->0) solver();
        out.flush();
        out.close();
    }

    static void solver(){
        int n = ri();
        int [][]a = new int[3][n];
        for(int i=0; i<n; ++i) a[0][i] = ri();
        for(int i=0; i<n; ++i) a[1][i] = ri();
        for(int i=0; i<n; ++i) a[2][i] = ri();
        ArrayList<Integer> ans = new ArrayList<>();
        ans.add(a[0][0]);
        for(int i=1; i<n-1; ++i){
            if(a[0][i] != ans.get(ans.size()-1)) ans.add(a[0][i]);
            else if(a[1][i] != ans.get(ans.size()-1)) ans.add(a[1][i]);
            else if(a[2][i] != ans.get(ans.size()-1)) ans.add(a[2][i]);
        }
        if(a[0][n-1] != ans.get(0) && a[0][n-1] != ans.get(ans.size()-1)) ans.add(a[0][n-1]);
        else if(a[1][n-1] != ans.get(0) && a[1][n-1] != ans.get(ans.size()-1)) ans.add(a[1][n-1]);
        else if(a[2][n-1] != ans.get(0) && a[2][n-1] != ans.get(ans.size()-1)) ans.add(a[2][n-1]);
        for(int x:ans) out.print(x + " ");
        out.println();
    }

    static class FastScanner {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer("");

        String next() {
            while (!st.hasMoreTokens())
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            return st.nextToken();
        }

        int nextInt() {
            return parseInt(next());
        }

        int[] readArray(int n) {
            int[] a = new int[n];
            for (int i = 0; i < n; i++) a[i] = nextInt();
            return a;
        }

        long nextLong() {
            return parseLong(next());
        }

        double nextDouble() {
            return parseDouble(next());
        }
    }

    static final int IBIG = 1000000007;
    static final int IMAX = 2147483647;
    static final int IMIN = -2147483648;
    static final long LMAX = 9223372036854775807L;
    static final long LMIN = -9223372036854775808L;

    static int ri() {
        return in.nextInt();
    }

    static double rd() {
        return in.nextDouble();
    }

    static long rl() {
        return in.nextLong();
    }

    static String rs() {
        return in.next();
    }

    static double[] rda(int n) {
        double[] a = new double[n];
        for (int i = 0; i < n; ++i) a[i] = in.nextDouble();
        return a;
    }

    static int[] ria(int n) {
        int[] a = new int[n];
        for (int i = 0; i < n; ++i) a[i] = in.nextInt();
        return a;
    }

    static long[] rla(int n) {
        long[] a = new long[n];
        for (int i = 0; i < n; ++i) a[i] = in.nextLong();
        return a;
    }

    static char[] rca() {
        return in.next().toCharArray();
    }

    static long powmod(long a, long b, long mod) {
        long res = 1;
        while (b > 0) {
            if ((b & 1) != 0) res *= a;
            a *= a;
            res %= mod;
            a %= mod;
            b = b >> 2;
        }
        return res;
    }

    static long powmod(int a, int b, long mod) {
        long res = 1;
        long a_copy = a;
        long b_copy = b;
        while (b_copy > 0) {
            if ((b_copy & 1) != 0) res *= a_copy;
            a_copy *= a_copy;
            res %= mod;
            a_copy %= mod;
            b_copy = b_copy >> 2;
        }
        return res;
    }

    static void pr(int n) {
        out.print(n);
    }

    static void pr(long n) {
        out.print(n);
    }

    static void pr(String n) {
        out.print(n);
    }

    static void pr(char n) {
        out.print(n);
    }

    static void pr(char[] a) {
        out.print(a.toString());
    }

    static void prln(int n) {
        out.println(n);
    }

    static void prln(long n) {
        out.println(n);
    }

    static void prln(String n) {
        out.println(n);
    }

    static void prln(char n) {
        out.println(n);
    }

    static void prln(char[] a) {
        out.println(a.toString());
    }

    static int[] copy(int[] a) {
        int[] b = new int[a.length];
        for (int i = 0; i < a.length; ++i) {
            b[i] = a[i];
        }
        return b;
    }

    static long[] copy(long[] a) {
        long[] b = new long[a.length];
        for (int i = 0; i < a.length; ++i) {
            b[i] = a[i];
        }
        return b;
    }

    static char[] copy(char[] a) {
        char[] b = new char[a.length];
        for (int i = 0; i < a.length; ++i) {
            b[i] = a[i];
        }
        return b;
    }

    static int[] reverse(int[] a) {
        int[] b = new int[a.length];
        for (int i = 0; i < a.length; ++i) {
            b[a.length - 1 - i] = a[i];
        }
        return b;
    }

    static long[] reverse(long[] a) {
        long[] b = new long[a.length];
        for (int i = 0; i < a.length; ++i) {
            b[a.length - 1 - i] = a[i];
        }
        return b;
    }

    static char[] reverse(char[] a) {
        char[] b = new char[a.length];
        for (int i = 0; i < a.length; ++i) {
            b[a.length - 1 - i] = a[i];
        }
        return b;
    }
}