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.util.*;



public class LEET1 {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int T = scanner.nextInt();
        for (int a = 0; a < T; a++) {
            int n = scanner.nextInt();
            int[] arr1 = new int[n];
            int[] arr2 = new int[n];
            int[] arr3 = new int[n];


            for (int i = 0; i < n; i++) {
                arr1[i] = scanner.nextInt();
            }
            for (int i = 0; i < n; i++) {
                arr2[i] = scanner.nextInt();
            }
            for (int i = 0; i < n; i++) {
                arr3[i] = scanner.nextInt();
            }
            f1(n,0,-1,arr1,arr2,arr3,new LinkedList<>());
        }
    }
    public static boolean f1(int n,int i,int prev,int[] arr1,int[] arr2,int[] arr3,LinkedList<Integer> list) {

        boolean m = false,k=false,l=false;
        if (i == n) {
            System.out.println(list);
            return true;
        }
        if (prev != -1) {
            if (i!=n-1) {
                if (arr1[i] != prev) {
                    LinkedList<Integer> l1 = new LinkedList<>(list);
                    l1.add(arr1[i]);
                    m = f1(n, i + 1, arr1[i], arr1, arr2, arr3, l1);
                }
                if (!m && arr2[i] != prev) {
                    LinkedList<Integer> l2 = new LinkedList<>(list);
                    l2.add(arr2[i]);
                    k = f1(n, i + 1, arr2[i], arr1, arr2, arr3, l2);
                } else return m;
                if (!k && arr3[i] != prev) {
                    LinkedList<Integer> l3 = new LinkedList<>(list);
                    l3.add(arr3[i]);
                    l = f1(n, i + 1, arr3[i], arr1, arr2, arr3, l3);
                    if (l) return true;
                    else return false;
                } else return k;
            }
            else {
                if (arr1[i] != prev&&arr1[i]!=list.get(0)) {
                    LinkedList<Integer> l1 = new LinkedList<>(list);
                    l1.add(arr1[i]);
                    m = f1(n, i + 1, arr1[i], arr1, arr2, arr3, l1);
                }
                if ((!m||arr1[i]==list.get(0)) && arr2[i] != prev&&arr2[i]!=list.get(0)) {
                    LinkedList<Integer> l2 = new LinkedList<>(list);
                    l2.add(arr2[i]);
                    k = f1(n, i + 1, arr2[i], arr1, arr2, arr3, l2);
                } else return m;
                if ((!k||arr2[i]==list.get(0)) && arr3[i] != prev&&arr3[i]!=list.get(0)) {
                    LinkedList<Integer> l3 = new LinkedList<>(list);
                    l3.add(arr3[i]);
                    l = f1(n, i + 1, arr3[i], arr1, arr2, arr3, l3);
                    if (l) return true;
                    else return false;
                } else return k;
            }
        } else {
            LinkedList<Integer> l1 = new LinkedList<>(list);
            l1.add(arr1[i]);
            m = f1(n, i + 1, arr1[i], arr1, arr2, arr3, l1);
            if (!m) {
                LinkedList<Integer> l2 = new LinkedList<>(list);
                l2.add(arr2[i]);
                k = f1(n, i + 1, arr2[i], arr1, arr2, arr3, l2);
            }else return m;
            if (!k) {
                LinkedList<Integer> l3 = new LinkedList<>(list);
                l3.add(arr3[i]);
                l= f1(n, i + 1, arr3[i], arr1, arr2, arr3, l3);
                if (l) return true;
                else return false;
            } else return k;

        }

    }

}