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 = int(input())
for i in range(t):
    n = int(input())
    a = list(map(int, input().split()))
    b = list(map(int, input().split()))
    c = list(map(int, input().split()))
    p = []
    # f = []
    f = []

    if a[0] != a[1] or a[0] != b[1] or a[0] != c[1]:
        if a[0] != a[n-1] or a[0] != b[n-1] or a[0] != c[n-1]:
            f.append(a[0])
    elif b[0] != a[1] or b[0] != b[1] or b[0] != c[1]:
        if a[0] != a[n-1] or a[0] != b[n-1] or a[0] != c[n-1]:
            f.append(b[0])
    elif c[0] != a[1] or c[0] != b[1] or c[0] != c[1]:
        if c[0] != a[n-1] or c[0] != b[n-1] or c[0] != c[n-1]:
            f.append(c[0])
    val = f[0]
    for j in range(1, n):
        if j == n-1:
            if a[j] != f[j-1] and a[j] != val:
                if a[j] != a[(j+1)%n] or a[j] != b[(j+1)%n] or a[j] != c[(j+1)%n]:
                    f.append(a[j])
                    continue
            if b[j] != f[j-1] and b[j] != val:
                if b[j] != a[(j+1)%n] or b[j] != b[(j+1)%n] or b[j] != c[(j+1)%n]:
                    f.append(b[j])
                    continue
            if c[j] != f[j-1] and c[j] != val:
                if c[j] != a[(j+1)%n] or c[j] != b[(j+1)%n] or c[j] != c[(j+1)%n]:
                    f.append(c[j])
                    continue
        elif a[j] != f[j-1]:
            if a[j] != f[j-1] and a[j] != val:
                if a[j] != a[j+1] or a[j] != b[j+1] or a[j] != c[j+1]:
                    f.append(a[j])
                    continue
        if b[j] != f[j-1]:
            if b[j] != f[j-1] and b[j] != val:
                if b[j] != a[j+1] or b[j] != b[j+1] or b[j] != c[j+1]:
                    f.append(b[j])
                    continue

        if c[j] != f[j-1]:
            if c[j] != f[j-1] and c[j] != val:
                if c[j] != a[j+1] or c[j] != b[j+1] or c[j] != c[j+1]:
                    f.append(c[j])
                    continue

    print(*f)