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 S
{
    public static void main(String args[])
    {
        Scanner scan=new Scanner(System.in);
        int test=scan.nextInt();
        while(test>0)
        {
            test--;
            int n=scan.nextInt();
            int a[]=new int[n];
            for(int i=0;i<n;i++)
            a[i]=scan.nextInt();
            int b[]=new int[n];
            for(int i=0;i<n;i++)
            b[i]=scan.nextInt();
            int c[]=new int[n];
            for(int i=0;i<n;i++)
            c[i]=scan.nextInt();
            int ans[]=new int[n];
            boolean check=true;
            int j=0;
            int present[]={a[j],b[j],c[j]};
            int current=present[0];
            ans[j]=current;
            int index=1;
            j++;
            for(int i=1;i<n;i++)
            {
                int next[]={a[i],b[i],c[i]};
                index=0;
                check=true;
                while(check)
                {
                  if(next[0]!=current)
                  {
                      current=next[0];
                      ans[j]=current;
                      check=false;
                      break;
                  }
                  else if(next[1]!=current)
                  {
                      current=next[1];
                      ans[j]=current;
                      check=false;
                      break;
                  }
                  else if(next[2]!=current)
                  {
                      current=next[2];
                      ans[j]=current;
                      check=false;
                      break;
                  }
                  else
                  {
                      if(index<3)
                      current=present[index];
                      else
                      {
                          current=present[0];
                          index=0;
                      }
                     
                      index++;
                      check=true;ans[j-1]=current;
                  }
                }
                present=next;
                j++;
            }
            //check the first and last element ;
            int f=ans[0];
            int l=ans[n-1];
            if(f==l&&n>2)
            {
             int sc=ans[n-2];
             if(present[0]!=f&&present[0]!=sc)
             ans[n-1]=present[0];
             else if(present[1]!=f&&present[1]!=sc)
             ans[n-1]=present[1];
             else
             ans[n-1]=present[2];
             
             
            }
            for(int k:ans)
            System.out.print(k+" ");
            System.out.println();
        }
       
    }
}