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;
namespace forever {
template <class type1, class type2>
inline istream &operator>>(istream &in, pair<type1, type2> &p) {
  in >> p.first >> p.second;
  return in;
}
template <class type1, class type2>
inline ostream &operator<<(ostream &o, pair<type1, type2> &p) {
  o << "(" << p.first << "," << p.second << ") ";
  return o;
}
template <class type>
inline ostream &operator<<(ostream &o, vector<type> &v) {
  for (long long i = 0; i < v.size(); i++) o << v[i] << " ";
  return o;
}
template <class type>
inline ostream &operator<<(ostream &o, set<type> &v) {
  for (long long x : v) o << x << " ";
  return o;
}
template <class t>
inline istream &operator>>(istream &in, vector<t> &p) {
  for (auto &x : p) in >> x;
  return in;
}
long long power(long long x, long long y, long long p = 1000000007) {
  long long res = 1;
  x %= p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y >>= 1;
    x = (x * x) % p;
  }
  return res;
}
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(pair<uint64_t, uint64_t> x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x.first + FIXED_RANDOM) ^
           (splitmix64(x.second + FIXED_RANDOM) >> 1);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
}  // namespace forever
using namespace forever;
void test_cases() {
  long long n;
  cin >> n;
  vector<long long> a(n), b(n), c(n);
  cin >> a >> b >> c;
  vector<long long> ans(n);
  ans[0] = a[0];
  for (long long i = 1; i < n - 1; i++) {
    if (ans[i - 1] == a[i])
      ans[i] = b[i];
    else if (ans[i - 1] == b[i])
      ans[i] = c[i];
    else
      ans[i] = b[i];
  }
  if (ans[0] != a[n - 1] && ans[n - 2] != a[n - 1])
    ans[n - 1] = a[n - 1];
  else if (ans[0] != b[n - 1] && ans[n - 2] != b[n - 1])
    ans[n - 1] = b[n - 1];
  else if (ans[0] != c[n - 1] && ans[n - 2] != c[n - 1])
    ans[n - 1] = c[n - 1];
  cout << ans << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t = 1;
  cin >> t;
  for (long long i = 1; i <= t; i++) {
    test_cases();
  }
  return 0;
}
