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.
t=int(input())
for _ in range(0,t):
    n=int(input())
    ls1=list(map(int,input().split()))
    ls2=list(map(int,input().split()))
    ls3=list(map(int,input().split()))
    p=[]

    prev=-999
    for i in range(0,n-1):
        if ls1[i]!=prev:
            prev=ls1[i]
            print(ls1[i],end=" ")
            p.append(ls1[i])
        elif ls2[i]!=prev:
            prev=ls2[i]
            print(ls2[i],end=" ")
            p.append(ls2[i])
        else:
            prev=ls3[i]
            print(ls3[i],end=" ")
            p.append(ls3[i])
    if ls1[-1]!=p[0] and ls1[-1]!=prev:
        print(ls1[-1])
    elif ls2[-1]!=p[0] and ls2[-1]!=prev:
        print(ls2[-1])
    elif ls3[-1]!=p[0] and ls3[-1]!=prev:
        print(ls3[-1])
    # print()





    # ls=[]
    # ls.extend(ls1)
    # ls.extend(ls2)
    # # print(ls)
    # ls.sort()
    # prev=-999
    # i=-1
    # for ii in range(0,n):
    #     rev=0
    #     if ls1[ii]==prev:
    #         rev=1
    #     if rev==0:
    #         if i==n-1:
    #             break
    #         i+=1

    #         print(ls1[ii],end=" ")

    #         if i==n-1:
    #             break
    #         i+=1

    #         print(ls2[ii],end=" ")
    #         if i==n-1:
    #             break
    #         i+=1

    #         print(ls3[ii],end=" ")
    #         prev=ls3[ii]
    #     else:
    #         if i==n-1:
    #             break
    #         i+=1

    #         print(ls3[ii],end=" ")

    #         if i==n-1:
    #             break
    #         i+=1

    #         print(ls2[ii],end=" ")
    #         if i==n-1:
    #             break
    #         i+=1

    #         print(ls1[ii],end=" ")
    #         prev=ls1[ii]


        




    # # counter = 0
    # # for i in range(0,n):
    # #     if f==0:
    # #         print(ls[counter],end=" ")
    # #         f=1
    # #     else:
    # #         print(ls[counter*-1-1],end=" ")
    # #         counter+=1
    # #         f=0
    # # print()
    # # prev=-9999999
    # # for i in range(0,n,3):
    # #     index=int(i/3)
    # #     if ls1[i]!=prev:
    # #         print(ls1[index],end=" ")
    # #         print(ls2[index],end=" ")
    # #         print(ls3[index],end=" ")
    # #         prev=ls3[index]
    # #     else:
    # #         print(ls3[index],end=" ")
    # #         print(ls2[index],end=" ")
    # #         print(ls1[index],end=" ")
    # #         prev=ls1[index]
        
    # # print()            
    
    # # # count=3
    
    # # # prev=0
    # # # this=1
    # # # while count<n:
    # # #     count+=1
    # # #     if ls1[]
        