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;
struct num {
  string x;
  string y;
};
long long pow(long long base, long long os) {
  if (os == 0) return 1;
  long long z = pow(base, os / 2);
  z *= z;
  return (os % 2 ? z * base : z);
}
bool isprime(int x) {
  for (long long i = 2; i * i <= x; i++) {
    if (x % i == 0) return false;
  }
  return true;
}
int prime[1000010] = {};
void sieve(int n) {
  for (int i = 2; i * i <= n; i++) {
    if (prime[i] == 0)
      for (int j = i * i; j <= n; j += i) {
        prime[j] = 1;
      }
  }
  for (int i = 2; i <= n; i++)
    if (prime[i] ^ 1) cout << (i);
}
long long a[5050], b[5050], c[5050];
void run_case() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++) cin >> b[i];
  for (int i = 0; i < n; i++) cin >> c[i];
  long long ans[5050];
  ans[0] = a[0];
  for (int i = 1; i < n - 1; i++) {
    if (a[i] != ans[i - 1]) {
      ans[i] = a[i];
      continue;
    }
    if (b[i] != ans[i - 1]) {
      ans[i] = b[i];
      continue;
    }
    if (c[i] != ans[i - 1]) {
      ans[i] = c[i];
      continue;
    }
  }
  if (a[n - 1] != ans[n - 2] && a[n - 1] != ans[0])
    ans[n - 1] = a[n - 1];
  else if (b[n - 1] != ans[n - 2] && b[n - 1] != ans[0])
    ans[n - 1] = b[n - 1];
  else
    ans[n - 1] = c[n - 1];
  for (int i = 0; i < n; i++) cout << ans[i] << " \n"[i == n - 1];
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int tests = 1;
  cin >> tests;
  while (tests--) run_case();
}