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.math.BigInteger;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashMap;
import java.util.HashSet;
import java.util.List;
import java.util.PriorityQueue;
import java.util.Random;
import java.util.StringTokenizer;

public class A {

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

		FastScanner sc = new FastScanner();
		int t = sc.nextInt();
		first: while (t-- != 0) {

			int n = sc.nextInt();
			int a[] = sc.readArray(n);
			int b[] = sc.readArray(n);
			int c[] = sc.readArray(n);
			ArrayList<Integer> lis = new ArrayList<Integer>();
			lis.add(c[0]);
			int pre = c[0];
			for(int i=1; i<n; i++) {
				if(pre == a[i]) {
					lis.add(b[i]);
					pre = b[i];
				}
				else {
					lis.add(a[i]);
					pre = a[i];
				}
			}
			pre = lis.get(0);
			int pre2 = lis.get(lis.size()-2);
			int val = lis.get(n-1);
			if(pre == val) {
				if(pre != a[n-1] && pre2 != a[n-1]) {
					lis.set(n-1, a[n-1]);
				}
				else if(pre != b[n-1] && pre2 != b[n-1]) {
					lis.set(n-1, b[n-1]);
				}
				else  {
					lis.set(n-1, c[n-1]);
				}
				
			}
			
			
			
			for(int e : lis)
				System.out.print(e+" ");
			System.out.println();
			

		}

	}

	static void shuffleSort(long[] a) {
		Random get = new Random();
		for (int i = 0; i < a.length; i++) {
			int r = get.nextInt(a.length);
			long temp = a[i];
			a[i] = a[r];
			a[r] = temp;
		}
	}

	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 Integer.parseInt(next());
		}

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

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

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

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

		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return str;
		}
	}

	static int ceil(int x, int y) {
		return (x % y == 0 ? x / y : (x / y + 1));
	}

	static long ceil(long x, long y) {
		return (x % y == 0 ? x / y : (x / y + 1));
	}

	static int max(int x, int y) {
		return Math.max(x, y);
	}

	static int min(int x, int y) {
		return Math.min(x, y);
	}

	static long max(long x, long y) {
		return Math.max(x, y);
	}

	static long min(long x, long y) {
		return Math.min(x, y);
	}

	static int min(int a[]) {
		int x = 1_000_000_00_9;
		for (int i = 0; i < a.length; i++)
			x = min(x, a[i]);
		return x;
	}

	static int max(int a[]) {
		int x = -1_000_000_00_9;
		for (int i = 0; i < a.length; i++)
			x = max(x, a[i]);
		return x;
	}

	static long min(long a[]) {
		long x = (long) 3e18;
		for (int i = 0; i < a.length; i++)
			x = min(x, a[i]);
		return x;
	}

	static long max(long a[]) {
		long x = -(long) 3e18;
		for (int i = 0; i < a.length; i++)
			x = max(x, a[i]);
		return x;
	}

	static int power(int x, int y) {
		int res = 1;
		while (y > 0) {
			if (y % 2 == 1)
				res = (res * x);
			y >>= 1;
			x = (x * x);
		}
		return res;
	}

	static long power(long x, long y) {
		long res = 1;
		while (y > 0) {
			if (y % 2 == 1)
				res = (res * x);
			y >>= 1;
			x = (x * x);
		}
		return res;
	}

	static long power(long x, long y, long mod) {
		long res = 1;
		x %= mod;
		while (y > 0) {
			if (y % 2 == 1)
				res = (res * x) % mod;
			y >>= 1;
			x = (x * x) % mod;
		}
		return res;
	}

	static void intsort(int[] a) {
		List<Integer> temp = new ArrayList<Integer>();
		for (int i = 0; i < a.length; i++)
			temp.add(a[i]);
		Collections.sort(temp);
		for (int i = 0; i < a.length; i++)
			a[i] = temp.get(i);
	}

	static void longsort(long[] a) {
		List<Long> temp = new ArrayList<Long>();
		for (int i = 0; i < a.length; i++)
			temp.add(a[i]);
		Collections.sort(temp);
		for (int i = 0; i < a.length; i++)
			a[i] = temp.get(i);
	}

	static void reverseintsort(int[] a) {
		List<Integer> temp = new ArrayList<Integer>();
		for (int i = 0; i < a.length; i++)
			temp.add(a[i]);
		Collections.sort(temp);
		Collections.reverseOrder();
		for (int i = 0; i < a.length; i++)
			a[i] = temp.get(i);
	}

	static void reverselongsort(long[] a) {
		List<Long> temp = new ArrayList<Long>();
		for (int i = 0; i < a.length; i++)
			temp.add(a[i]);
		Collections.sort(temp);
		Collections.reverseOrder();
		for (int i = 0; i < a.length; i++)
			a[i] = temp.get(i);
	}

	static class longpair implements Comparable<longpair> {
		long x, y;

		longpair(long x, long y) {
			this.x = x;
			this.y = y;
		}

		public int compareTo(longpair p) {
			return Long.compare(this.x, p.x);
		}
	}

	static class intpair implements Comparable<intpair> {
		int x, y;

		intpair(int x, int y) {
			this.x = x;
			this.y = y;
		}

		@Override
		public int compareTo(intpair o) {
			return Integer.compare(this.x, o.x);
		}

		//		a = new pair [n];
		//		a[i] = new pair(coo,cost);

	}

	public static int gcd(int a, int b) {
		BigInteger b1 = BigInteger.valueOf(a);
		BigInteger b2 = BigInteger.valueOf(b);
		BigInteger gcd = b1.gcd(b2);
		return gcd.intValue();
	}

	public static long gcd(long a, long b) {
		BigInteger b1 = BigInteger.valueOf(a);
		BigInteger b2 = BigInteger.valueOf(b);
		BigInteger gcd = b1.gcd(b2);
		return gcd.longValue();
	}
}