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.
#!/usr/bin/env pypy
from __future__ import division, print_function
from collections import defaultdict, Counter, deque
from future_builtins import ascii, filter, hex, map, oct, zip
from itertools import imap as map, izip as zip, permutations, combinations, combinations_with_replacement,product
from __builtin__ import xrange as range
from math import ceil, factorial, log,tan,pi,cos,sin,radians
from _continuation import continulet
from cStringIO import StringIO
from io import IOBase
import __pypy__
from bisect import bisect, insort, bisect_left, bisect_right
from fractions import Fraction
from functools import reduce
from decimal import *
import copy
import string
import sys
import os
import re
inf = float('inf')
mod_ = int(1e9) + 7
mod = 998244353

def factors(n):
    from functools import reduce
    return set(reduce(list.__add__, ([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0)))

def sieve(n):
    arr=[1]*n
    for i in range(2,int(n**0.5)+1):
        if arr[i]==1:
            for j in range(i,n,i):
                arr[j]=i
    return arr


def valid(arr):
    return arr[0][1]==arr[1][0]

def main():
    for _ in range(int(input())):
        n=int(input())
        arr=list(map(int,input().split()))
        barr=list(map(int,input().split()))
        carr=list(map(int,input().split()))
        ans1=[arr[0]]
        for i in range(1,n):
            if ans1[-1]==arr[i]:
                ans1.append(barr[i])
            else:
                ans1.append(arr[i])
        print(*ans1)















    
    










# region fastio

BUFSIZE = 8192

class FastI(IOBase):
    def __init__(self, file):
        self._fd = file.fileno()
        self._buffer = StringIO()
        self.newlines = 0

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count("\n") + (not b)
            ptr = self._buffer.tell()
            self._buffer.seek(0, 2), self._buffer.write(b), self._buffer.seek(ptr)
        self.newlines -= 1
        return self._buffer.readline()


class FastO(IOBase):
    def __init__(self, file):
        self._fd = file.fileno()
        self._buffer = __pypy__.builders.StringBuilder()
        self.write = lambda s: self._buffer.append(s)

    def flush(self):
        os.write(self._fd, self._buffer.build())
        self._buffer = __pypy__.builders.StringBuilder()


def print(*args, **kwargs):
    sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
    at_start = True
    for x in args:
        if not at_start:
            file.write(sep)
        file.write(str(x))
        at_start = False
    file.write(kwargs.pop("end", "\n"))
    if kwargs.pop("flush", False):
        file.flush()

def gcd(x, y):
    while y:
        x, y = y, x % y
    return x

sys.stdin, sys.stdout = FastI(sys.stdin), FastO(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")

# endregion

if __name__ == "__main__":
    main()