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 PYTHON3):

Solution:

import sys
import bisect as bi
import math
from collections import defaultdict as dd
import heapq
import itertools
import io
import os
import operator
input=sys.stdin.readline
##input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
import random
##sys.setrecursionlimit(10**7)
##fo=open("output2.txt","w")
##fi=open("input2.txt","w")
mo=10**9+7
def cin():
    return map(int,sin().split())
def ain():            
    return list(map(int,sin().split()))
def sin():
    return input()
def inin():
    return int(input())

for _ in range(inin()):
    n=inin()
    a=ain();b=ain();c=ain()
    prev=a[0];ans=[prev]
    for i in range(1,n):
        l=[a[i],b[i],c[i]]
        for j in l:
            if(j!=prev):
                if(i==n-1):
                    if(j!=ans[0]):
                        ans+=[j]
                        break
                else:
                    ans+=[j]
                    prev=j
                    break
    print(*ans)
                
        
    
    

        
    

    
        
    
##def msb(n):n|=n>>1;n|=n>>2;n|=n>>4;n|=n>>8;n|=n>>16;n|=n>>32;n|=n>>64;return n-(n>>1) #2 ki power
##def pref(a,n,f):             
##    pre=[0]*n
##    if(f==0):         ##from beginning
##        pre[0]=a[0]
##        for i in range(1,n):
##            pre[i]=a[i]+pre[i-1]
##    else:              ##from end
##        pre[-1]=a[-1]
##        for i in range(n-2,-1,-1):
##            pre[i]=pre[i+1]+a[i]
##    return pre
##maxint=10**24 
##def kadane(a,size): 
##    max_so_far = -maxint - 1
##    max_ending_here = 0
##       
##    for i in range(0, size): 
##        max_ending_here = max_ending_here + a[i] 
##        if (max_so_far < max_ending_here): 
##            max_so_far = max_ending_here 
##  
##        if max_ending_here < 0: 
##            max_ending_here = 0   
##    return max_so_far
##def power(x, y):
##    if(y == 0):return 1
##    temp = power(x, int(y / 2))%mo
##    if (y % 2 == 0):return (temp * temp)%mo 
##    else:
##        if(y > 0):return (x * temp * temp)%mo 
##        else:return ((temp * temp)//x )%mo