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.*;
import java.util.*;

/**
    @author KhanhNguyenn
 */
 
public class D{
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        Task solver = new Task();
        solver.solve(in, out);
        out.close();
    }
    // main solver
    static class Task{
        public void solve(InputReader in, PrintWriter out) {
            int t= in.nextInt();
            while(t-->0){
                int n= in.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]= in.nextInt();
                for(int i=0;i<n;i++) b[i]= in.nextInt();
                for(int i=0;i<n;i++) c[i]= in.nextInt();
                int first= a[0];
                int pre= a[0];
                out.print(a[0]+" ");
                for(int i=1;i<n-1;i++){
                    int choose= -1;
                    if(a[i]!=pre) choose=a[i];
                    else if(b[i]!=pre) choose=b[i];
                    else choose= c[i];
                    out.print(choose+" ");
                    pre=choose;
                }
                if(a[n-1]!=pre && a[n-1]!=first) out.println(a[n-1]);
                else if(b[n-1]!=pre && b[n-1]!=first) out.println(b[n-1]);
                else if(c[n-1]!=pre && c[n-1]!=first) out.println(c[n-1]);
            }

            
        }

        public int[][] lcs(char[] A, char[] B) {
            int n= A.length, m= B.length;
            int[][] dp= new int[n+1][m+1];

            for(int i=1;i<=n;i++) {
                for(int j=1;j<=m;j++) {
                    if(A[i-1]==B[j-1]){
                        dp[i][j]= dp[i-1][j-1]+1;
                    }
                    else {
                        dp[i][j]= Math.max(dp[i-1][j],dp[i][j-1]);
                    }
                }
            }
            return dp;
        }

        public List<Pair> dfs(int[][] dp, char[] A, char[] B) {
            int n= A.length, m= B.length;
            int i= n, j=m;
            List<Pair> l= new ArrayList<>();
            while(dp[i][j]!=0) {
                if(A[i-1]==B[j-1]) {
                    l.add(new Pair(i-1,j-1));
                    i--;j--;
                }
                else{
                    if(dp[i-1][j]>dp[i][j-1]) i--;
                    else j--;
                }
            }
            return l;
        }
    }
    static class Pair implements Comparable<Pair>{
        public int x, y;
        public Pair(int x, int y){
            this.x = x;
            this.y = y;
        }
 
        @Override
        public int compareTo(Pair o) {
            if (this.x > o.x){
                return 1;
            }
            else if (this.x < o.x){
                return -1;
            }
            else{
                return Integer.compare(this.y, o.y);
            }
        }
    }
    // fast input reader class;
    static class InputReader {
        BufferedReader br;
        StringTokenizer st;
 
        public InputReader(InputStream stream) {
            br = new BufferedReader(new InputStreamReader(stream));
        }
 
        public String nextToken() {
            while (st == null || !st.hasMoreTokens()) {
                String line = null;
                try {
                    line = br.readLine();
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
                if (line == null) {
                    return null;
                }
                st = new StringTokenizer(line);
            }
            return st.nextToken();
        }
 
        public int nextInt() {
            return Integer.parseInt(nextToken());
        }
        public double nextDouble(){
            return Double.parseDouble(nextToken());
        }
        public long nextLong(){
            return Long.parseLong(nextToken());
        }
    }
}