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:
from sys import stdin, stdout, setrecursionlimit
#import threading

# tail-recursion optimization
# In case of tail-recusion optimized code, have to use python compiler.
# Otherwise, memory limit may exceed.
# declare the class Tail_Recursion_Optimization
class Tail_Recursion_Optimization:
    def __init__(self, RECURSION_LIMIT, STACK_SIZE):
        setrecursionlimit(RECURSION_LIMIT)
        threading.stack_size(STACK_SIZE)
        return None
        
class SOLVE:
    def solve(self):
        R = stdin.readline
        #f = open('input.txt');R = f.readline
        W = stdout.write
        
        ans = []
        
        for i in range(int(R())):
            n = int(R())
            a = [int(x) for x in R().split()]
            b = [int(x) for x in R().split()]
            c = [int(x) for x in R().split()]
            
            tmp = []
            for j in range(n):
                if j == 0:
                    tmp.append(a[0])
                
                elif j == n-1:
                    if a[j] != tmp[0] and a[j] != tmp[n-2]:
                        tmp.append(a[j])
                    elif b[j] != tmp[0] and b[j] != tmp[n-2]:
                        tmp.append(b[j])
                    elif c[j] != tmp[0] and c[j] != tmp[n-2]:
                        tmp.append(c[j])
                        
                else:
                    if tmp[j-1] == a[j]:
                        if tmp[j-1] == b[j]:
                            tmp.append(c[j])
                        elif tmp[j-1] == c[j]:
                            tmp.append(b[j])
                        else:
                            tmp.append(b[j])
                    
                    elif tmp[j-1] == b[j]:
                        if tmp[j-1] == c[j]:
                            tmp.append(a[j])
                        elif tmp[j-1] == a[j]:
                            tmp.append(c[j])
                        else:
                            tmp.append(c[j])
                    
                    elif tmp[j-1] == c[j]:
                        if tmp[j-1] == a[j]:
                            tmp.append(b[j])
                        elif tmp[j-1] == b[j]:
                            tmp.append(a[j])
                        else:
                            tmp.append(a[j])
                    else:
                        tmp.append(a[j])
            print(*tmp)
                
        return 0
        
def main():
    s = SOLVE()
    s.solve()
#Tail_Recursion_Optimization(10**7, 100*1024**2) # recursion-call size, stack-size in byte (MB*1024**2)
#threading.Thread(target=main).start()
main()