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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
struct elem {
  int first, second, third;
};
void sort_vector(vector<elem>& arr, int N) {
  for (int i = 0; i < N; ++i) {
    int a, b, c;
    a = b = c = 999999999;
    if (arr[i].first < a) a = arr[i].first;
    if (arr[i].second < a) a = arr[i].second;
    if (arr[i].third < a) a = arr[i].third;
    if (arr[i].first < b && a < arr[i].first) b = arr[i].first;
    if (arr[i].second < b && a < arr[i].second) b = arr[i].second;
    if (arr[i].third < b && a < arr[i].third) b = arr[i].third;
    if (arr[i].first > a && arr[i].first > b) c = arr[i].first;
    if (arr[i].second > a && arr[i].second > b) c = arr[i].second;
    if (arr[i].third > a && arr[i].third > b) c = arr[i].third;
    arr[i].first = a, arr[i].second = b, arr[i].third = c;
  }
}
void testcase() {
  int N;
  cin >> N;
  vector<elem> arr(N);
  for (int i = 0; i < N; ++i) cin >> arr[i].first;
  for (int i = 0; i < N; ++i) cin >> arr[i].second;
  for (int i = 0; i < N; ++i) cin >> arr[i].third;
  sort_vector(arr, N);
  int first_taken_value = -1;
  vector<int> final_p(N);
  for (int i = 0; i < N; ++i) {
    if (first_taken_value == -1) {
      first_taken_value = arr[i].first;
      final_p[i] = arr[i].first;
    } else {
      if (i == N - 1) {
        if (arr[i].first != final_p[i - 1] &&
            arr[i].first != first_taken_value) {
          final_p[i] = arr[i].first;
        } else if (arr[i].second != final_p[i - 1] &&
                   arr[i].second != first_taken_value) {
          final_p[i] = arr[i].second;
        } else
          final_p[i] = arr[i].third;
      } else {
        if (arr[i].first != final_p[i - 1]) {
          final_p[i] = arr[i].first;
        } else if (arr[i].second != final_p[i - 1]) {
          final_p[i] = arr[i].second;
        } else
          final_p[i] = arr[i].third;
      }
    }
  }
  for (int i = 0; i < N; ++i) cout << final_p[i] << " ";
  cout << "\n";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) testcase();
}