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 spf[1004];
void sieve() {
  spf[1] = 1;
  for (long long i = 2; i < 1004; i++) spf[i] = i;
  for (long long i = 4; i < 1004; i += 2) spf[i] = 2;
  for (long long i = 3; i * i < 1004; i++) {
    if (spf[i] == i) {
      for (long long j = i * i; j < 1004; j += i)
        if (spf[j] == j) spf[j] = i;
    }
  }
}
vector<long long> getFactorization(long long x) {
  vector<long long> ret;
  while (x != 1) {
    ret.push_back(spf[x]);
    x = x / spf[x];
  }
  return ret;
}
long long gcd(long long a, long long b) {
  if (b == 0) {
    return a;
  }
  return gcd(b, a % b);
}
long long __gcd(long long a, long long b) {
  if (b > a) {
    swap(a, b);
  }
  return gcd(a, b);
}
long long lcm(long long a, long long b) {
  long long lcm1 = (a / __gcd(a, b)) * b;
  return lcm1;
}
long long power(long long x, long long y) {
  long long res = 1;
  while (y > 0) {
    if (y & 1) {
      res = res * x;
    }
    y = y >> 1;
    x = x * x;
  }
  return res;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t = 1;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    long long a[n];
    long long b[n];
    long long 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];
    }
    cout << a[0] << " ";
    long long prev = a[0];
    long long i = 1;
    while (i < n - 1) {
      if (prev == a[i]) {
        cout << b[i] << " ";
        prev = b[i];
      } else {
        cout << a[i] << " ";
        prev = a[i];
      }
      i++;
    }
    if (a[i] != prev && a[i] != a[0]) {
      cout << a[i];
    } else if (b[i] != prev && b[i] != a[0]) {
      cout << b[i];
    } else {
      cout << c[i];
    }
    cout << "\n";
  }
  return 0;
}