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.
# Design_by_JOKER

from math import *
from cmath import *
from itertools import *
from decimal import *  # su dung voi so thuc
from fractions import *  # su dung voi phan so
from sys import *

# getcontext().prec = x # lay x-1 chu so sau giay phay ( thuoc decimal)
# Decimal('12.3') la 12.3 nhung Decimal(12.3) la 12.30000000012
# Fraction(a) # tra ra phan so bang a (Fraction('1.23') la 123/100 Fraction(1.23) la so khac (thuoc Fraction)
# a = complex(c, d) a = c + d(i) (c = a.real, d = a.imag)
# a.capitalize() bien ki tu dau cua a(string) thanh chu hoa, a.lower() bien a thanh chu thuong, tuong tu voi a.upper()
# a.swapcase() doi nguoc hoa thuong, a.title() bien chu hoa sau dau cach, a.replace('a', 'b', slg)
# a.join['a', 'b', 'c'] = 'a'a'b'a'c, a.strip('a') bo dau va cuoi ki tu 'a'(rstrip, lstrip)
# a.split('a', slg = -1) cat theo ki tu 'a' slg lan(rsplit(), lsplit()), a.count('aa', dau = 0, cuoi= len(a)) dem slg
# a.startswith('a', dau = 0, cuoi = len(a)) co bat dau bang 'a' ko(tuong tu endswith())
# a.index("aa") vi tri dau tien xuat hien (rfind())
# input = open(".inp", mode='r')  a = input.readline()
# out = open(".out", mode='w')# a.index(val)

for _ in range(int(input())):
    n = int(input())
    a = [int(x) for x in input().split()]
    b = [int(x) for x in input().split()]
    c = [int(x) for x in input().split()]
    p = a
    for x in range(1, n-1):
        if p[x] == p[x-1]:
            p[x] = b[x]
    if p[n-1] == p[n-2] or p[n-1] == p[0]:
        p[n-1] = b[n-1]
    if p[n - 1] == p[n - 2] or p[n - 1] == p[0]:
        p[n-1] = c[n-1]
    for x in range(n):
        print(p[x], end = ' ')
    print()