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;
template <class T>
using minheap = priority_queue<T, vector<T>, greater<T>>;
template <typename T>
void setmax(T& a, T b) {
  a = max(a, b);
};
template <typename T>
void setmin(T& a, T b) {
  a = min(a, b);
};
template <typename T>
bool in(T lo, T v, T hi) {
  return lo <= v && v <= hi;
};
template <long long MOD>
struct ModInt {
  long long x = 0;
  ModInt() { x = 0; }
  ModInt(long long _x) {
    x = _x;
    if (x >= MOD || x <= -MOD) x %= MOD;
    if (x < 0) x += MOD;
  }
  friend ModInt operator+(ModInt a, ModInt b) { return ModInt(a.x + b.x); }
  friend ModInt operator-(ModInt a, ModInt b) { return ModInt(a.x - b.x); }
  friend ModInt operator*(ModInt a, ModInt b) { return ModInt(a.x * b.x); }
  friend ModInt operator/(ModInt a, ModInt b) { return a * inv(b); }
  friend ModInt power(ModInt a, long long b) {
    if (b == 0) return 1;
    ModInt c = power(a, b / 2);
    return c * c * ((b & 1) ? a : 1);
  }
  friend ModInt inv(ModInt a) { return power(a, MOD - 2); }
  friend ostream& operator<<(ostream& os, const ModInt a) {
    os << a.x;
    return os;
  }
};
const int mod = 1e9 + 7;
using mint = ModInt<mod>;
struct Solution {
  void solve() {
    int n;
    cin >> n;
    vector<int> a(n), b(n), c(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];
    vector<int> d(n);
    d[0] = a[0];
    for (int i = 1; i < n; i++) {
      if (a[i] != d[i - 1])
        d[i] = a[i];
      else if (b[i] != d[i - 1])
        d[i] = b[i];
      else
        d[i] = c[i];
    }
    if (d[0] == d[n - 1]) {
      if (a[n - 1] != d[0] && a[n - 1] != d[n - 2])
        d[n - 1] = a[n - 1];
      else if (b[n - 1] != d[0] && b[n - 1] != d[n - 2])
        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';
  }
};
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    Solution second;
    second.solve();
  }
  return 0;
}