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:
'''
      ___           ___                         ___                       ___           ___                         ___
     /\__\         /\  \         _____         /\  \                     /\  \         /\  \                       /\__\
    /:/ _/_        \:\  \       /::\  \        \:\  \       ___         /::\  \       |::\  \         ___         /:/ _/_
   /:/ /\  \        \:\  \     /:/\:\  \        \:\  \     /\__\       /:/\:\__\      |:|:\  \       /\__\       /:/ /\  \
  /:/ /::\  \   ___  \:\  \   /:/  \:\__\   ___ /::\  \   /:/__/      /:/ /:/  /    __|:|\:\  \     /:/  /      /:/ /::\  \
 /:/_/:/\:\__\ /\  \  \:\__\ /:/__/ \:|__| /\  /:/\:\__\ /::\  \     /:/_/:/__/___ /::::|_\:\__\   /:/__/      /:/_/:/\:\__\
 \:\/:/ /:/  / \:\  \ /:/  / \:\  \ /:/  / \:\/:/  \/__/ \/\:\  \__  \:\/:::::/  / \:\~~\  \/__/  /::\  \      \:\/:/ /:/  /
  \::/ /:/  /   \:\  /:/  /   \:\  /:/  /   \::/__/       ~~\:\/\__\  \::/~~/~~~~   \:\  \       /:/\:\  \      \::/ /:/  /
   \/_/:/  /     \:\/:/  /     \:\/:/  /     \:\  \          \::/  /   \:\~~\        \:\  \      \/__\:\  \      \/_/:/  /
     /:/  /       \::/  /       \::/  /       \:\__\         /:/  /     \:\__\        \:\__\          \:\__\       /:/  /
     \/__/         \/__/         \/__/         \/__/         \/__/       \/__/         \/__/           \/__/       \/__/

'''
"""
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"""
import sys
import math
import collections
import operator as op
from collections import deque
from math import gcd, inf, sqrt, pi, cos, sin, ceil, log2, floor
from bisect import bisect_right, bisect_left, bisect

# sys.stdin = open('input.txt', 'r')
# sys.stdout = open('output.txt', 'w')

from functools import reduce
from sys import stdin, stdout, setrecursionlimit
setrecursionlimit(2**20)


def ncr(n, r):
    r = min(r, n - r)
    numer = reduce(op.mul, range(n, n - r, -1), 1)
    denom = reduce(op.mul, range(1, r + 1), 1)
    return numer // denom  # or / in Python 2


def prime_factors(n):
    i = 2
    factors = []
    while i * i <= n:
        if n % i:
            i += 1
        else:
            n //= i
            factors.append(i)
    if n > 1:
        factors.append(n)
    return (list(factors))


def sumDigits(no):
    return 0 if no == 0 else int(no % 10) + sumDigits(int(no / 10))

MOD = 1000000007
PMOD = 998244353
N = 10**9 + 7
LOGN = 30
T = 1
T = int(stdin.readline())
for _ in range(T):
    # n, k = list(map(int, stdin.readline().rstrip().split()))
    n = int(stdin.readline())
    a = list(map(int, stdin.readline().rstrip().split()))
    b = list(map(int, stdin.readline().rstrip().split()))
    c = list(map(int, stdin.readline().rstrip().split()))
    # a = str(stdin.readline().strip('\n'))
    # b = list(stdin.readline().strip('\n'))
    # m = int(stdin.readline())
    # s = list(map(int, stdin.readline().rstrip().split())\
    if n == 1:
        print(*a)
    for i in range(1, n):
        if a[i] == a[i - 1]:
            if a[i] != b[i]:
                a[i] = b[i]
            else:
                a[i] = c[i]
    if a[0] == a[-1]:
        if a[-1] != b[-1] and b[-1] != a[-2]:
            a[-1] = b[-1]
        elif a[-1] != c[-1] and c[-1] != a[-2]:
            a[-1] = c[-1]
    print(*a)