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.*;
import java.util.*;
import java.math.BigInteger;
public class Main {
	static PrintWriter out=new PrintWriter(System.out);
	
	public static void main(String[] args) throws Exception {
		Scanner sc = new Scanner(System.in);
		int t=sc.nextInt();
		while(t-->0) {
			int n=sc.nextInt();
			int a[]=sc.nextIntArr(n);
			int b[]=sc.nextIntArr(n);
			int c[]=sc.nextIntArr(n);
			int arr[]=new int[n];
			arr[0]=a[0];
			for(int i=1;i<n;i++) {
				if(a[i]==arr[i-1])arr[i]=b[i];
				else arr[i]=a[i];
			}
			if(arr[n-1]==arr[0]) {
				if(c[n-1]!=arr[n-2])arr[n-1]=c[n-1];
				else if(b[n-1]!=arr[n-2])arr[n-1]=b[n-1];
				else if(a[n-1]!=arr[n-2])arr[n-1]=a[n-1];
			}
			for(int i:arr)out.print(i+" ");
			out.println();
		}
		out.close();
	}
	
	static class Pair implements Comparable<Pair>{
		int x;
		int y;
		public Pair(int x,int y) {
			this.x=x;
			this.y=y;
		}
		public int compareTo(Pair o) {
			return (this.x!=o.x)?this.x-o.x:this.y-o.y;
		}
		public String toString() {
			return "("+this.x+","+this.y+")";
		}
		
	}
	static class Scanner {
		StringTokenizer st;
		BufferedReader br;
		public Scanner(InputStream s) {
			br = new BufferedReader(new InputStreamReader(s));
		}
		public Scanner(String file) throws FileNotFoundException {
			br = new BufferedReader(new FileReader(file));
		}
		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 String nextLine() throws IOException {
			return br.readLine();
		}
		public double nextDouble() throws IOException {
			return Double.parseDouble(next());
		}
		public boolean ready() throws IOException {
			return br.ready();
		}
		public int[] nextIntArr(int n) throws IOException{
			int arr[]=new int[n];
			for(int i=0;i<n;i++)arr[i]=nextInt();
			return arr;
		}
		public BigInteger BigInteger(String s) throws IOException{
			BigInteger a=new BigInteger(s);
			return a;
		}
	}
}