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.
// Added -> dsu, mp, ncr, pair, sieve, suffix, zalgo
// To add -> Geometry, Advance DS Topics, Articulation Points & Bridges, Tries, LCA, Segment Tree

import java.io.*;
import java.util.*;
 
public class utkarsh {

    public static void main(String[] args) {
        new utkarsh().run();
    }
    
    BufferedReader br;
    PrintWriter out;
    
    long mod = (long) (1e9 + 7), inf = (long) (3e18);

    void solve() {
        int t = ni();
        while(t-- > 0) {
            int n = ni();
            int a[] = na(n);
            int b[] = na(n);
            int c[] = na(n);
            int pre = -1, st = 0;
            for(int i = 0; i < n-1; i++) {
                if(a[i] != pre) pre = a[i];
                else if(b[i] != pre)    pre = b[i];
                else pre = c[i];
                out.print(pre +" ");
                if(i == 0)  st = pre;
            }
            if(a[n-1] != pre && a[n-1] != st)   out.println(a[n-1]);
            else if(b[n-1] != pre && b[n-1] != st)   out.println(b[n-1]);
            else if(c[n-1] != pre && c[n-1] != st)   out.println(c[n-1]);
        }
    }
    
    // -------- I/O Template -------------
    
    char nc() {
        return ns().charAt(0);
    }
    
    String nLine() {
        try {
            return br.readLine();
        } catch(IOException e) {
            return "-1";
        }
    }
    
    double nd() {
        return Double.parseDouble(ns());
    }
    
    long nl() {
        return Long.parseLong(ns());
    }
    
    int ni() {
        return Integer.parseInt(ns());
    }

    int[] na(int n) {
        int a[] = new int[n];
        for(int i = 0; i < n; i++)  a[i] = ni();
        return a;
    }
    
    StringTokenizer ip;
    
    String ns() {
        if(ip == null || !ip.hasMoreTokens()) {
            try {
                ip = new StringTokenizer(br.readLine());
            } catch(IOException e) {
                throw new InputMismatchException();
            }
        }
        return ip.nextToken();
    }

    void run() {
        try {
            if (System.getProperty("ONLINE_JUDGE") == null) {
                br = new BufferedReader(new FileReader("input.txt"));
                out = new PrintWriter("output.txt");
            } else {
                br = new BufferedReader(new InputStreamReader(System.in));
                out = new PrintWriter(System.out);
            }
        } catch (FileNotFoundException e) {
            System.out.println(e);
        }
        solve();
        out.flush();
    }
}