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.
import sys
import math
 
# Constants
mod = 998244353
 
 
# Functions Settings
dynamic_factorial = False
 
 
 
# Pre-defined function
# Begin
 
# Testcase handler
def main(test_case = False):
    n = int(input()) if test_case else 1
    for _ in range(n):
        test()
        print()
 
# Binary method of power
def fast_pow(a:int, b:int):
    res = 1
    while b > 0:
        if b & 1:
            res *= a
        a *= a
        b >>= 1
    return res
 
# Combination of 2
def c2(n):
    return n * (n-1) // 2
 
# Get the list form of input
def get():
    return list(map(int, input().split()))
 
# Number of 1s in binary form
def bits(n: int):
    return list(bin(n)).count('1')
 
# For interactive problems
def flush():
    sys.stdout.flush()
 
# Print array and split it with space
def parr(arr):
    print(*arr, sep=' ')
 
# GCD (int a, int b)
# Return: value of GCD(a, b)
def gcd(a, b):
    while b:
        if b % a == 0:
            break
        tmp = a
        a = b % a
        b = tmp
    return a
 
# Extended GCD (int a, int b)
# Return: array of [GCD(a, b), [x0, y0]]
# Where a * x0 + b * y0 = 1
# If GCD(a, b) != 1 then return False
def ext_gcd(a: int, b: int):
    if (b == 0):
        return [a, [1, 0]]
 
    res = ext_gcd(b, a % b)
    g = res[0]
    if g != 1:
        return False
    x1 = res[1][0]
    y1 = res[1][1]
    x = y1
    y = x1 - y1 * (a // b)
 
    return [g, [x, y]]
 
# Get all different values of array a with their count
def get_set(a):
    res = {}
    for x in a:
        if x in res:
            res[x] += 1
        else:
            res[x] = 1
    return res
 
# Factorial
fact = {0: 1}
def factorial(n):
    if dynamic_factorial:
        if n not in fact:
            fact[n] = factorial(n-1) * n
        return fact[n]
    else:
        res = 1
        while n > 0:
            res = (res * n) % mod
            n -= 1
        return res
 
# Sum of array
def sum_of(a):
    res = 0
    for x in a:
        res = (res + x) % mod
    return res
 
# Prod of array
def prod_of(a):
    res = 1
    for x in a:
        res = (res * x) % mod
    return res
 
# End
 
def test():
    n = get()[0]
    a, b, c = get(), get(), get()
    ans = [a[0]]
    i = 1
    while i < n-1:
        if ans[i-1] == a[i]:
            ans.append(b[i])
        else:
            ans.append(a[i])
        i += 1
    ans.append(0)
    if a[i] != ans[i-1] and a[i] != ans[0]:
        ans[i] = a[i]
    elif b[i] != ans[i-1] and b[i] != ans[0]:
        ans[i] = b[i]
    elif c[i] != ans[i-1] and c[i] != ans[0]:
        ans[i] = c[i]
    
    parr(ans)

main(test_case = True)