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:
T_ON = 1
DEBUG_ON = 1
MOD = 998244353


def solve():
    n = read_int()
    A = read_ints()
    B = read_ints()
    C = read_ints()
    D = A[:]
    for i in range(1, n):
        if D[i] == D[i-1]:
            D[i] = B[i]

    if D[-1] == D[-2] or D[-1] == D[0]:
        D[-1] = B[-1]
    if D[-1] == D[-2] or D[-1] == D[0]:
        D[-1] = C[-1]
    print_nums(D)


def main():
    T = read_int() if T_ON else 1
    for i in range(T):
        solve()


def debug(*xargs):
    if DEBUG_ON:
        print(*xargs)


from collections import *
import math


#---------------------------------FAST_IO---------------------------------------

import os
import sys
from io import BytesIO, IOBase
# region fastio

BUFSIZE = 8192


class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    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(b"\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()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)


class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")


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

#----------------------------------IO_WRAP--------------------------------------

def read_int():
    return int(input())


def read_ints():
    return list(map(int, input().split()))


def print_nums(nums):
    print(" ".join(map(str, nums)))


def YES():
    print("YES")


def Yes():
    print("Yes")


def NO():
    print("NO")


def No():
    print("No")


def First():
    print("First")


def Second():
    print("Second")


#----------------------------------FIB--------------------------------------

def fib(n):
    """
    the nth fib, start from zero
    """
    a, b = 0, 1
    for _ in range(n):
        a, b = b, a + b
    return a


def fib_ns(n):
    """
    the first n fibs, start from zero
    """
    assert n >= 1
    f = [0 for _ in range(n + 1)]
    f[0] = 0
    f[1] = 1
    for i in range(2, n + 1):
        f[i] = f[i - 1] + f[i - 2]
    return f


def fib_to_n(n):
    """
    return fibs <= n, start from zero

    n=8
    f=[0,1,1,2,3,5,8]
    """
    f = []
    a, b = 0, 1
    while a <= n:
        f.append(a)
        a, b = b, a + b
    return f

#----------------------------------MOD--------------------------------------

def gcd(a, b):
    if a == 0:
        return b
    return gcd(b % a, a)


def xgcd(a, b):
    """return (g, x, y) such that a*x + b*y = g = gcd(a, b)"""
    x0, x1, y0, y1 = 0, 1, 1, 0
    while a != 0:
        (q, a), b = divmod(b, a), a
        y0, y1 = y1, y0 - q * y1
        x0, x1 = x1, x0 - q * x1
    return b, x0, y0


def lcm(a, b):
    d = gcd(a, b)
    return a * b // d


def is_even(x):
    return x % 2 == 0


def is_odd(x):
    return x % 2 == 1


def modinv(a, m):
    """return x such that (a * x) % m == 1"""
    g, x, _ = xgcd(a, m)
    if g != 1:
        raise Exception('gcd(a, m) != 1')
    return x % m


def mod_add(x, y):
    x += y
    while x >= MOD:
        x -= MOD
    while x < 0:
        x += MOD
    return x


def mod_mul(x, y):
    return (x * y) % MOD


def mod_pow(x, y):
    if y == 0:
        return 1
    if y % 2:
        return mod_mul(x, mod_pow(x, y - 1))
    p = mod_pow(x, y // 2)
    return mod_mul(p, p)


def mod_inv(y):
    return mod_pow(y, MOD - 2)


def mod_div(x, y):
    # y^(-1): Fermat little theorem, MOD is a prime
    return mod_mul(x, mod_inv(y))

#---------------------------------PRIME---------------------------------------

def is_prime(n):
    if n == 1:
        return False
    for i in range(2, int(n ** 0.5) + 1):
        if n % i:
            return False
    return True


def gen_primes(n):
    """
    generate primes of [1..n] using sieve's method
    """
    P = [True for _ in range(n + 1)]
    P[0] = P[1] = False
    for i in range(int(n ** 0.5) + 1):
        if P[i]:
            for j in range(2 * i, n + 1, i):
                P[j] = False
    return P

#---------------------------------MAIN---------------------------------------

main()