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;
namespace fastExpress {
template <typename T>
inline T read() {
  T f = 1, x = 0;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return x * f;
}
template <typename T>
inline void write(T x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
template <typename T>
inline void print_char(T x, char ch) {
  write(x), putchar(ch);
}
template <typename T>
inline void char_print(char ch, T x) {
  putchar(ch);
  write(x);
}
template <typename T>
inline void print_space(T x) {
  print_char(x, ' ');
}
template <typename T>
inline void space_print(T x) {
  char_print(' ', x);
}
template <typename T>
inline void println(T x) {
  print_char(x, '\n');
}
template <typename T>
inline void lnprint(T x) {
  char_print('\n', x);
}
template <typename T>
inline void print_array_in_range1(T* a, int l, int r) {
  for (int(i) = (l); (i) <= (r); ++(i)) print_char(a[i], " \n"[i == r]);
}
template <typename T>
inline void print_array_in_range2(T* a, int l, int r) {
  println(r - l + 1), print_array_in_range1(a, l, r);
}
template <typename T>
inline void print_array_in_range3(T* a, int l, int r) {
  for (int(i) = (r); (i) >= (l); --(i)) print_char(a[i], " \n"[i == l]);
}
template <typename T>
inline void print_array_in_range4(T* a, int l, int r) {
  println(r - l + 1), print_array_in_range3(a, l, r);
}
template <typename T>
inline void print_array_in_range5(T* a, int l, int r) {
  for (int(i) = (l); (i) <= (r); ++(i)) println(a[i]);
}
template <typename T>
inline void print_array_in_range6(T* a, int l, int r) {
  println(r - l + 1), print_array_in_range5(a, l, r);
}
template <typename T>
inline void print_array_in_range7(T* a, int l, int r) {
  for (int(i) = (r); (i) >= (l); --(i)) println(a[i]);
}
template <typename T>
inline void print_array_in_range8(T* a, int l, int r) {
  println(r - l + 1), print_array_in_range7(a, l, r);
}
template <typename T>
inline void print_array_in_range9(T* a, int l, int r) {
  print_space(r - l + 1), print_array_in_range1(a, l, r);
}
template <typename T>
inline void print_array_in_range10(T* a, int l, int r) {
  print_space(r - l + 1), print_array_in_range3(a, l, r);
}
template <typename T>
inline void print_array_in_range11(T* a, int l, int r) {
  print_space(r - l + 1), print_array_in_range5(a, l, r);
}
template <typename T>
inline void print_array_in_range12(T* a, int l, int r) {
  print_space(r - l + 1), print_array_in_range7(a, l, r);
}
template <typename T>
inline void print_array_in_range13(T* a, int l, int r, int n) {
  println(n), print_array_in_range1(a, l, r);
}
template <typename T>
inline void print_array_in_range14(T* a, int l, int r, int n) {
  println(n), print_array_in_range1(a, l, r);
}
template <typename T>
inline void print_array_in_range15(T* a, int l, int r, int n) {
  println(n), print_array_in_range1(a, l, r);
}
template <typename T>
inline void print_array_in_range16(T* a, int l, int r, int n) {
  println(n), print_array_in_range1(a, l, r);
}
template <typename T>
inline void print_array_in_range17(T* a, int l, int r, int n) {
  print_space(n), print_array_in_range1(a, l, r);
}
template <typename T>
inline void print_array_in_range18(T* a, int l, int r, int n) {
  print_space(n), print_array_in_range1(a, l, r);
}
template <typename T>
inline void print_array_in_range19(T* a, int l, int r, int n) {
  print_space(n), print_array_in_range1(a, l, r);
}
template <typename T>
inline void print_array_in_range20(T* a, int l, int r, int n) {
  print_space(n), print_array_in_range1(a, l, r);
}
template <typename T>
inline void print_array1(T* a, int n) {
  print_array_in_range1(a, 1, n);
}
template <typename T>
inline void print_array2(T* a, int n) {
  println(n), print_array1(a, n);
}
template <typename T>
inline void print_array3(T* a, int n) {
  print_array_in_range3(a, 1, n);
}
template <typename T>
inline void print_array4(T* a, int n) {
  println(n), print_array3(a, n);
}
template <typename T>
inline void print_array5(T* a, int n) {
  print_array_in_range5(a, 1, n);
}
template <typename T>
inline void print_array6(T* a, int n) {
  println(n), print_array5(a, n);
}
template <typename T>
inline void print_array7(T* a, int n) {
  print_array_in_range7(a, 1, n);
}
template <typename T>
inline void print_array8(T* a, int n) {
  println(n), print_array7(a, n);
}
template <typename T>
inline void print_array9(T* a, int n) {
  print_space(n), print_array1(a, n);
}
template <typename T>
inline void print_array10(T* a, int n) {
  print_space(n), print_array3(a, n);
}
template <typename T>
inline void print_array11(T* a, int n) {
  print_space(n), print_array5(a, n);
}
template <typename T>
inline void print_array12(T* a, int n) {
  print_space(n), print_array7(a, n);
}
template <typename T>
inline void vector_print(vector<T> a, int l, int r) {
  for (int(i) = (l); (i) <= (r); ++(i)) print_char(a[i], " \n"[i == r]);
}
template <typename T>
inline void map_print(map<T, T> a, int l, int r) {
  for (int(i) = (l); (i) <= (r); ++(i)) print_char(a[i], " \n"[i == r]);
}
template <typename T>
inline void stack_print(stack<T> a, char ch) {
  while (!a.empty()) print_char(a.top(), ch), a.pop();
}
template <typename T>
inline void queue_print(queue<T> a, char ch) {
  while (!a.empty()) print_char(a.front(), ch), a.pop();
}
template <typename T>
inline void pq_print(priority_queue<T> a, char ch) {
  while (!a.empty()) print_char(a.top(), ch), a.pop();
}
inline void print_string_in_range(string s, int l, int r) {
  for (int(i) = (l); (i) <= (r); ++(i)) putchar(s[i]);
}
inline void print_chars_in_range(char* a, int l, int r) {
  for (int(i) = (l); (i) <= (r); ++(i)) putchar(a[i]);
}
inline void print_allstring(string s) {
  print_string_in_range(s, 0, (int)s.size() - 1);
}
inline void filein(string s) { freopen((s + ".in").c_str(), "r", stdin); }
inline void fileout(string s) { freopen((s + ".out").c_str(), "w", stdout); }
inline void file(string s) { filein(s), fileout(s); }
}  // namespace fastExpress
using namespace fastExpress;
int n, a[107], b[107], c[107], ans[107];
int main() {
  int Test = read<int>();
  while (Test--) {
    n = read<int>(), memset((ans), (0), sizeof((ans)));
    for (int(i) = (1); (i) <= (n); ++(i)) a[i] = read<int>();
    for (int(i) = (1); (i) <= (n); ++(i)) b[i] = read<int>();
    for (int(i) = (1); (i) <= (n); ++(i)) c[i] = read<int>();
    for (int(i) = (1); (i) <= (n); ++(i)) {
      if (i == 1)
        ans[i] = a[i];
      else if (i != n) {
        if (ans[i - 1] != a[i])
          ans[i] = a[i];
        else if (ans[i - 1] != b[i])
          ans[i] = b[i];
        else
          ans[i] = c[i];
      } else {
        if (ans[1] != a[n] && ans[n - 1] != a[n])
          ans[n] = a[n];
        else if (ans[1] != b[n] && ans[n - 1] != b[n])
          ans[n] = b[n];
        else if (ans[1] != c[n] && ans[n - 1] != c[n])
          ans[n] = c[n];
      }
    }
    print_array1(ans, n);
  }
  return 0;
}