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 int digits(long long int n) { return floor(log10(double(n))) + 1; }
void faster() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
}
long long int modularExponentiation(long long int x, long long int n,
                                    long long int M) {
  long long int result = 1;
  while (n > 0) {
    if (n % 2 == 1) {
      result = ((result % M) * (x % M)) % M;
    }
    x = ((x % M) * (x % M)) % M;
    n = n / 2;
  }
  return result % M;
}
long long int d, x, y;
void extendedEuclid(long long int A, long long int B) {
  if (B == 0) {
    d = A;
    x = 1;
    y = 0;
  } else {
    extendedEuclid(B, A % B);
    int temp = x;
    x = y;
    y = temp - (A / B) * y;
  }
}
long long int nCrModp(long long int n, long long int r) {
  if (r > n - r) r = n - r;
  long long int C[r + 1];
  memset(C, 0, sizeof(C));
  C[0] = 1;
  for (long long int i = 1; i <= n; i++) {
    for (long long int j = min(i, r); j > 0; j--) {
      C[j] = (C[j] + C[j - 1]);
    }
  }
  return C[r];
}
long long int modInverse(long long int A, long long int M) {
  extendedEuclid(A, M);
  return (x % M + M) % M;
}
int main() {
  long long int t;
  cin >> t;
  while (t--) {
    long long int n;
    cin >> n;
    long long int ans[n];
    long long int arr[n];
    long long int brr[n];
    long long int crr[n];
    for (long long int i = 0; i < n; i++) {
      cin >> arr[i];
    }
    for (long long int i = 0; i < n; i++) {
      cin >> brr[i];
    }
    for (long long int i = 0; i < n; i++) {
      cin >> crr[i];
    }
    ans[0] = arr[0];
    cout << ans[0] << " ";
    for (long long int i = 1; i < n - 1; i++) {
      if (arr[i] != ans[i - 1])
        ans[i] = arr[i];
      else if (brr[i] != ans[i - 1])
        ans[i] = brr[i];
      else
        ans[i] = crr[i];
      cout << ans[i] << " ";
    }
    if (arr[n - 1] != ans[n - 2] && ans[0] != arr[n - 1])
      cout << arr[n - 1];
    else if (brr[n - 1] != ans[n - 2] && ans[0] != brr[n - 1])
      cout << brr[n - 1];
    else
      cout << crr[n - 1];
    cout << endl;
  }
}