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;
const long long MOD = 1000000007;
void solve();
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    solve();
    cout << "\n";
  }
  cerr << "time taken : " << (float)clock() / CLOCKS_PER_SEC << " secs" << endl;
  return 0;
}
long long ceils(long long x, long long y) { return x / y + ((x % y) != 0); }
long long gcd(long long a, long long b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
void solve() {
  long long n;
  cin >> n;
  long long a[n], b[n], c[n];
  for (long long i = 0; i < n; i++) cin >> a[i];
  for (long long i = 0; i < n; i++) cin >> b[i];
  for (long long i = 0; i < n; i++) cin >> c[i];
  long long ans[n];
  ans[0] = a[0];
  for (long long i = 1; i < n; i++) {
    ans[i] = a[i];
    if (ans[i] == ans[i - 1]) ans[i] = b[i];
    if (ans[i] == ans[i - 1]) ans[i] = c[i];
  }
  long long ok = 0;
  for (long long i = 1; i < n; i++) {
    if (ans[i] == ans[i - 1]) {
      ok = 1;
      break;
    }
  }
  if (ok) {
    ans[0] = b[0];
    for (long long i = 1; i < n; i++) {
      ans[i] = b[i];
      if (ans[i] == ans[i - 1]) ans[i] = a[i];
      if (ans[i] == ans[i - 1]) ans[i] = c[i];
    }
    ok = 0;
    for (long long i = 1; i < n; i++) {
      if (ans[i] == ans[i - 1]) {
        ok = 1;
        break;
      }
    }
  }
  if (ok) {
    ans[0] = c[0];
    for (long long i = 1; i < n; i++) {
      ans[i] = c[i];
      if (ans[i] == ans[i - 1]) ans[i] = a[i];
      if (ans[i] == ans[i - 1]) ans[i] = b[i];
    }
    ok = 0;
    for (long long i = 1; i < n; i++) {
      if (ans[i] == ans[i - 1]) {
        ok = 1;
        break;
      }
    }
  }
  if (ans[n - 1] == ans[0]) {
    if (ans[n - 1] == a[n - 1]) {
      if (ans[n - 2] == b[n - 1])
        ans[n - 1] = c[n - 1];
      else
        ans[n - 1] = b[n - 1];
    } else if (ans[n - 1] == b[n - 1]) {
      if (ans[n - 2] == a[n - 1])
        ans[n - 1] = c[n - 1];
      else
        ans[n - 1] = a[n - 1];
    } else {
      if (ans[n - 2] == b[n - 1])
        ans[n - 1] = a[n - 1];
      else
        ans[n - 1] = b[n - 1];
    }
  }
  for (long long i = 0; i < n; i++) cout << ans[i] << " ";
}
