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.
//package GraknLabs;

import java.util.*;
import java.io.*;

public class A {
    static Scanner sc;
    static PrintWriter out;

    public static void main(String[] args) {
        sc = new Scanner(System.in);
        out = new PrintWriter(System.out);


        int t = 1;
        t = sc.nextInt();
        while (t-- > 0) {
            forEachTestCase();
        }


        sc.close();
        out.close();
    }

    static void forEachTestCase() {
        int n=sc.nextInt();
        int a[][]=new int[3][n];
        for(int i=0;i<3;i++){
            for(int j=0;j<n;j++){
                a[i][j]=sc.nextInt();
            }
        }
        int last=-1,first=-1;
        for (int i = 0; i <n ; i++) {
            for (int j = 0; j <3 ; j++) {
                if( i!=n-1 && last!=a[j][i]){
                    out.print(a[j][i]+" ");
                    if(i==0){
                        first=a[j][i];
                    }
                    last=a[j][i];
                    break;
                }
                else if(i==n-1 && last!=a[j][i] && first!=a[j][i]){
                    out.println(a[j][i]);
                    break;
                }
            }
        }
    }

    static long mod = 1000000007;

    static long add(long a, long b) {
        a %= mod;
        b %= mod;
        return (a + b) % mod;
    }

    static long mul(long a, long b) {
        a %= mod;
        b %= mod;
        return (a * b) % mod;
    }

    static long sub(long a, long b) {
        a %= mod;
        b %= mod;
        return (((a - b) % mod) + mod) % mod;
    }

    static long gcd(long a, long b) {
        if (b == 0) return a;
        return gcd(b, a % b);
    }

    static long modPow(long a, long b) {
        long ans = 1;
        while (b > 0) {
            if (b % 2 == 1) {
                ans = mul(ans, a);
            }
            a = mul(a, a);
            b >>= 1;
        }
        return ans;
    }

    static long modInv(long a) {
        return modPow(a, mod - 2);
    }
}
