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;
long long ceill(long long x, long long y) {
  if (x % y == 0) {
    return x / y;
  }
  return (x / y) + 1;
}
long long prime(long long n) {
  vector<long long> v;
  while (n % 2 == 0) {
    v.push_back(2);
    n /= 2;
  }
  for (long long i = 3; i < sqrt(n); i += 1) {
    while (n % i == 0) {
      v.push_back(i);
      n /= i;
    }
  }
  if (n > 2) {
    v.push_back(n);
  }
  return v.size();
}
int min_sum(int n) {
  if (n == 0)
    return 0;
  else if (n == 1)
    return 2;
  return (n + 1 + min_sum(n / 2) + min_sum(n - n / 2 - 1));
}
void printDivisors(long long n) {
  vector<long long> v;
  for (long long i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) {
      if (n / i == i) {
        v.push_back(i);
      } else {
        v.push_back(i);
        v.push_back(n / i);
      }
    }
  }
  cout << v.size() << "\n";
}
bool numis(string s) {
  for (int i = 0; i < s.length(); i++)
    if (isdigit(s[i]) == false) return false;
  return true;
}
int fact(int a[], int n) {
  if (a[0] == a[1] && a[0] != a[2]) {
    return 2;
  }
  if (a[0] != a[1] && a[1] == a[2]) {
    return 0;
  }
  if (a[0] == a[2] && a[1] != a[0]) {
    return 1;
  }
  for (long long i = 3; i < n; i += 1) {
    if (a[i] != a[i - 1]) {
      return i;
    }
  }
}
long long b_Search(long long *a, long long l, long long r, long long x) {
  if (l > r) {
    return -1;
  }
  int m = ((l + r) / 2);
  if (x >= a[m] && x < a[m + 1]) {
    return m;
  }
  if (x > a[m]) {
    return b_Search(a, m + 1, r, x);
  }
  return b_Search(a, l, m - 1, x);
}
int sub(string s, string s2) {
  int l1 = s.size();
  int l2 = s2.size();
  int l = l1 - l2;
  for (int i = 0; i <= l; i++) {
    int j;
    for (j = 0; j < l2; j++) {
      if (s[i + j] != s2[j]) {
        break;
      }
    }
    if (j == l2) {
      s.erase(i, 1);
      return sub(s, s2);
    }
  }
  return s.size();
}
pair<string, int> findEntryWithLargestValue(map<string, int> m) {
  pair<string, int> mm = make_pair("", 0);
  map<string, int>::iterator itr;
  for (itr = m.begin(); itr != m.end(); ++itr) {
    if (itr->second > mm.second) {
      mm = make_pair(itr->first, itr->second);
    }
  }
  return mm;
}
void body() {
  int n;
  cin >> n;
  int a[n], b[n], c[n], d[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];
  }
  d[0] = a[0];
  for (int i = 1; i < n - 1; i++) {
    if (d[i - 1] == a[i]) {
      d[i] = b[i];
    } else {
      d[i] = a[i];
    }
  }
  if (d[n - 2] != a[n - 1] && d[0] != a[n - 1]) {
    d[n - 1] = a[n - 1];
  } else if (d[n - 2] != b[n - 1] && d[0] != b[n - 1]) {
    d[n - 1] = b[n - 1];
  } else {
    d[n - 1] = c[n - 1];
  }
  for (int i = 0; i < n; i++) {
    cout << d[i] << " ";
  }
  cout << "\n";
}
template <typename T>
class Graph {
  map<T, list<T> > l;

 public:
  void addEdge(int x, int y) {
    l[x].push_back(y);
    l[y].push_back(x);
  }
  void bfs(T src) {
    map<T, int> vis;
    queue<T> q;
    q.push(src);
    vis[src] = 1;
    while (!q.empty()) {
      int node = q.front();
      q.pop();
      cout << node << " ";
      for (int nbr : l[node]) {
        if (!vis[nbr]) {
          q.push(nbr);
          vis[nbr] = 1;
        }
      }
    }
  }
};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int t;
  cin >> t;
  while (t--) {
    body();
  }
  return 0;
}