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.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.Scanner;
import java.util.StringTokenizer;

public class fib {
	//static int r;
	static int aa, bb;
	static String x;
	static int inp[];
	static int len ;
	static long[][][] dp;

	public static long dp(int ind, int count, int exp) {
		if(ind==len) {
			if (count==0)
				return 0;
			else 
				return (long) 1e9;
		}
		if (dp[ind][count][exp]!=(long) 1e9)
			return dp[ind][count][exp];
		if (inp[ind]==0) {
			if(count!=0) {
				
				long a = dp(ind+1,count+1,0)+bb;
				long b = dp(ind+1,0,1)+aa+bb;
				return dp[ind][count][exp] = Math.min(a, b);
				
			}
				
			
			
			return dp[ind][count][exp] = dp(ind+1,count,0);
				
		}else {
			if(exp==1)
			return dp[ind][count][exp] = dp(ind+1,0,1);
			else
				{long a= dp(ind+1,0,1)+aa;
				long b= dp(ind+1,count+1,0);
				return dp[ind][count][exp]= Math.min(a, b);
				}
		}
	}

	public static long gcd(long i, long j) {
		if (j == 0)
			return i;
		return gcd(j, i % j);
	}

	public static int gcd(int i, int j) {
		if (j == 0)
			return i;
		return gcd(j, i % j);
	}

	public static long pow(int a, int e) {
		long res = 1;
		while (e != 0) {
			if ((e & 1) == 1)
				res = res * a;
			a = a * a;
			e >>= 1;

		}
		return res;
	}

	public static long perms(long x, int k) {
		long powerd = pow(10, (x + "").length() - 1);
		long nine = pow(9, (x + "").length() - 1);

		long b = x / powerd;

		if (b == k)
			return b * nine;
		if (b > k) {
			if ((x + "").length() == 1)
				return b;
			return (b - 1) * nine + perms(x % powerd, k);
		}
		if ((x + "").length() == 1)
			return b + 1;
		return (b) * nine + perms(x % powerd, k);

	}

	public static long zerperms(long x, int k) {
		long powerd = pow(10, (x + "").length() - 1);
		long nine = pow(9, (x + "").length() - 1);

		long b = x / powerd;

		if ((x + "").length() == 1)
			return b;
		return b * nine + perms(x % powerd, 0);

	}

	static class Scanner {
		BufferedReader br;
		StringTokenizer st;

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

		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

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

		public long nextLong() throws IOException {
			return Long.parseLong(next());
		}

		public double nextDouble() throws IOException {
			return Double.parseDouble(next());
		}

		public int[] nextIntArr(int n) throws IOException {
			int[] arr = new int[n];
			for (int i = 0; i < n; i++) {
				arr[i] = Integer.parseInt(next());
			}
			return arr;
		}

		public boolean ready() {
			try {
				return br.ready();
			} catch (IOException e) {
				throw new RuntimeException(e);
			}

		}
	}

	public static void main(String[] args) throws InterruptedException, IOException {

		Scanner sc = new Scanner(System.in);
		PrintWriter pw = new PrintWriter(System.out);

		int 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 []res =new int[n];
			res [0]= a[0];
			char sel ='a';
			for(int i = 1;i<n-1;i++) {
				if (a[i]==res[i-1])
				{
					res[i]=b[i];
					
				}
				else 
					res[i]=a[i];
			}
			if(a[n-1]!=res[n-2]&&a[n-1]!=res[0]) {
				res[n-1]=a[n-1];
			}
			else
				if(b[n-1]!=res[n-2]&&b[n-1]!=res[0]) {
					res[n-1]=b[n-1];
				}
				else 
					res[n-1]=c[n-1];
			
				
			for(int i=0;i<n;i++)
				pw.print(res[i]+" ");
			pw.println();
		}
		// pw.println(perms(tmp, k)-1);

		pw.close();

	}

}
