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.util.Scanner;

public class Circle_Coloring {
    static boolean canbreak = false;
    public static void printa(int [] p, int n){
        for (int i = 0; i < n - 1; i++ ){
            System.out.print((p[i]));
            System.out.print(' ');
        }
        System.out.println(p[n-1]);
    }
    public static boolean check(int x,int y, int z, int k, int n){
        if (k == n-1) {
            return x != y && x != z;

        }else{
            return x !=y;
        }
    }
    public static void search(int [] a, int [] b, int [] c, int [] p, int k, int n){
        if (k >= n){
            printa(p, n);
            canbreak = true;
        } else {
           if (k == 0){
              p[k] = a[k];
              search(a,b,c,p,k+1,n);
              p[k] = 0;
               p[k] = b[k];
               search(a,b,c,p,k+1,n);
               p[k] = 0;
               p[k] = c[k];
               search(a,b,c,p,k+1,n);
               p[k] = 0;
           } else {
               if (!canbreak && check(a[k], p[k-1], p[0], k, n)){
                   p[k] = a[k];
                   search(a,b,c,p,k+1,n);
                   p[k] = 0;
               }
               if (!canbreak && check(b[k], p[k-1], p[0], k, n)){
                   p[k] = b[k];
                   search(a,b,c,p,k+1,n);
                   p[k] = 0;
               }
               if (!canbreak && check(c[k], p[k-1], p[0], k, n)){
                   p[k] = c[k];
                   search(a,b,c,p,k+1,n);
                   p[k] = 0;
               }
           }
        }
    }
    public static void main(String [] args){
        Scanner s = new Scanner(System.in);
        int t = s.nextInt();
        s.nextLine();
        for (int i = 0; i < t; i++){
            int n = s.nextInt();
            s.nextLine();
            int [] a = new int[n];
            int [] b = new int [n];
            int [] c = new int [n];
            int [] p = new int [n];
            int [] v = new int [n];
            int [] istaken = new int [n];
            for (int j = 0; j < n; j++) {
                a[j] = s.nextInt();
            }
            s.nextLine();
            for (int j = 0; j < n; j++) {
                b[j] = s.nextInt();
            }
            s.nextLine();
            for (int j = 0; j < n; j++) {
                c[j] = s.nextInt();
            }
            s.nextLine();
            canbreak = false;
            search(a,b,c,p,0,n);
        }
    }
}