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;
void solve() {
  long long int n{};
  cin >> n;
  vector<long long int> a(n);
  vector<long long int> b(n);
  vector<long long int> c(n);
  vector<long long int> p(n);
  for (long long int i = 0; i < n; i++) {
    cin >> a.at(i);
  }
  for (long long int i = 0; i < n; i++) {
    cin >> b.at(i);
  }
  for (long long int i = 0; i < n; i++) {
    cin >> c.at(i);
  }
  p.at(0) = a.at(0);
  for (long long int i = 1; i < n - 1; i++) {
    if (a.at(i) != p.at(i - 1)) {
      p.at(i) = a.at(i);
    } else if (b.at(i) != p.at(i - 1)) {
      p.at(i) = b.at(i);
    } else if (c.at(i) != p.at(i - 1)) {
      p.at(i) = c.at(i);
    }
  }
  if (a.at(n - 1) != p.at(n - 2) && a.at(n - 1) != p.at(0)) {
    p.at(n - 1) = a.at(n - 1);
  } else if (b.at(n - 1) != p.at(n - 2) && b.at(n - 1) != p.at(0)) {
    p.at(n - 1) = b.at(n - 1);
  } else if (c.at(n - 1) != p.at(n - 2) && c.at(n - 1) != p.at(0)) {
    p.at(n - 1) = c.at(n - 1);
  }
  for (long long int i = 0; i < n; i++) {
    cout << p.at(i) << " ";
  }
  cout << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t{};
  cin >> t;
  while (t--) solve();
  return 0;
}