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;
int gcd(int a, int b) {
  if (b > a) return gcd(b, a);
  return b == 0 ? a : gcd(b, a % b);
}
bool solve(vector<long long>& ans, vector<long long> a, vector<long long> b,
           vector<long long> c, long long n, long long x) {
  if (x == n) return true;
  if (x == 0) {
    ans.push_back(a[0]);
    if (solve(ans, a, b, c, n, x + 1)) {
      return true;
    } else {
      ans.pop_back();
    }
    ans.push_back(b[0]);
    if (solve(ans, a, b, c, n, x + 1)) {
      return true;
    } else {
      ans.pop_back();
    }
    ans.push_back(c[0]);
    if (solve(ans, a, b, c, n, x + 1)) {
      return true;
    }
  } else if (x != n - 1) {
    if (ans.back() != a[x]) {
      ans.push_back(a[x]);
      if (solve(ans, a, b, c, n, x + 1)) {
        return true;
      } else
        ans.pop_back();
    } else if (ans.back() != b[x]) {
      ans.push_back(b[x]);
      if (solve(ans, a, b, c, n, x + 1)) {
        return true;
      } else
        ans.pop_back();
    } else if (ans.back() != c[x]) {
      ans.push_back(c[x]);
      if (solve(ans, a, b, c, n, x + 1)) {
        return true;
      } else
        ans.pop_back();
    }
  } else {
    if (ans.back() != a[x] && ans[0] != a[x]) {
      ans.push_back(a[x]);
      if (solve(ans, a, b, c, n, x + 1)) {
        return true;
      } else
        ans.pop_back();
    } else if (ans.back() != b[x] && ans[0] != b[x]) {
      ans.push_back(b[x]);
      if (solve(ans, a, b, c, n, x + 1)) {
        return true;
      } else
        ans.pop_back();
    } else if (ans.back() != c[x] && ans[0] != c[x]) {
      ans.push_back(c[x]);
      if (solve(ans, a, b, c, n, x + 1)) {
        return true;
      } else
        ans.pop_back();
    }
  }
  return false;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    vector<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];
    if (n == 1) {
      cout << a[0] << endl;
      continue;
    }
    vector<long long> ans;
    if (solve(ans, a, b, c, n, 0)) {
      for (long long i = 0; i < n; i++) {
        cout << ans[i] << " ";
      }
    }
    cout << endl;
  }
  return 0;
}