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.
#include <bits/stdc++.h>
using namespace std;
template <typename T>
using min_heap = priority_queue<T, vector<T>, greater<T>>;
template <typename T>
istream& operator>>(istream& is, vector<T>& vec) {
  for (auto& v : vec) is >> v;
  return is;
}
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& v) {
  os << "[ ";
  for (int i = 0; i < v.size(); ++i) {
    os << v[i];
    if (i != v.size() - 1) os << ", ";
  }
  os << " ]\n";
  return os;
}
template <typename T>
ostream& operator<<(ostream& os, const set<T>& v) {
  os << "[ ";
  for (auto it : v) {
    os << it;
    if (it != *v.rbegin()) os << ", ";
  }
  os << " ]\n";
  return os;
}
template <typename T>
ostream& operator<<(ostream& os, const multiset<T>& v) {
  os << "[ ";
  for (auto it : v) os << it << ", ";
  os << " ]\n";
  return os;
}
template <typename T, typename S>
ostream& operator<<(ostream& os, const map<T, S>& v) {
  os << "{ ";
  for (auto it : v) os << it.first << " : " << it.second << ", ";
  os << " }\n";
  return os;
}
template <typename T, typename S>
ostream& operator<<(ostream& os, const pair<T, S>& v) {
  os << "(";
  os << v.first << ", " << v.second << ")";
  return os;
}
template <typename T>
ostream& operator<<(ostream& os, priority_queue<T> p) {
  os << "[ ";
  while (!p.empty()) {
    os << p.top() << " ,";
    p.pop();
  }
  os << " ]\n";
  return os;
}
inline long long pmod(long long i, long long n) { return (i % n + n) % n; }
const long long INF = 1e18;
const long long mod = 1e9 + 7;
void solve() {
  long long n;
  cin >> n;
  vector<vector<long long>> a(3, vector<long long>(n));
  cin >> a;
  long long prev = -1, te;
  for (long long(i) = (0); (i) <= (n - 2); ++(i)) {
    for (long long(j) = (0); (j) <= (2); ++(j)) {
      if (prev != a[j][i]) {
        prev = a[j][i];
        if (i == 0) te = prev;
        cout << prev << ' ';
        break;
      }
    }
  }
  for (long long(i) = (0); (i) <= (2); ++(i))
    if (a[i][n - 1] != prev && a[i][n - 1] != te) {
      cout << a[i][n - 1];
      return;
    }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  cin >> t;
  for (int t1 = 1; t1 <= t; ++t1) {
    solve();
    cout << '\n';
  }
}
