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.
#include <bits/stdc++.h>
using namespace std;
long long int mod_Expo(long long int x, long long int n, long long int M) {
  if (x == 0) return 0;
  if (n == 0)
    return 1;
  else if (n % 2 == 0)
    return mod_Expo((x * x) % M, n / 2, M);
  else
    return (x * mod_Expo((x * x) % M, (n - 1) / 2, M)) % M;
}
long long int binaryExponentiation(long long int x, long long int n) {
  if (x == 0) return 0;
  if (n == 0)
    return 1;
  else if (n % 2 == 0)
    return binaryExponentiation(x * x, n / 2);
  else
    return x * binaryExponentiation(x * x, (n - 1) / 2);
}
long long int GCD(long long int A, long long int B) {
  if (B == 0)
    return A;
  else
    return GCD(B, A % B);
}
bool prime_check(long long int x) {
  bool prime = (x >= 2);
  for (long long i = 2; i * i <= x; i++) {
    if (x % i == 0) {
      prime = false;
      break;
    }
  }
  return prime;
}
long long int min1(long long int a, long long int b, long long int c) {
  vector<long long int> v1, v2, v3;
  v1.push_back(b);
  v1.push_back(c);
  v1.push_back(a);
  sort(v1.begin(), v1.end());
  return v1[0];
}
long long int min(long long int a, long long int b) {
  if (a <= b)
    return a;
  else
    return b;
}
long long int bin_search(long long int x, long long int low, long long int high,
                         vector<long long int> v1) {
  while (high > low) {
    long long int mid = low + (high - low) / 2;
    if (v1[mid] > x)
      high = mid;
    else
      low = mid + 1;
  }
  return high;
}
long long int digitsum(long long int n) {
  long long int sum = 0;
  while (n != 0) {
    sum = sum + n % 10;
    n = n / 10;
  }
  return sum;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int n, m, i, j, t, f = 0, g = 0, x, y, z, k, a = 0, b = 0, c = 0, q,
                               x1, x2, y1, y2, x3, max2, min1 = 1000000, x4,
                               sum, d, M = 998244353, r, p, sum1 = 0, sum2 = 0,
                               sum3 = 0, sum4, sum5, max1 = 0, sum6;
  cin >> t;
  while (t--) {
    cin >> n;
    long long int arr1[n], arr2[n], arr3[n];
    for (i = 0; i < n; i++) cin >> arr1[i];
    for (i = 0; i < n; i++) cin >> arr2[i];
    for (i = 0; i < n; i++) cin >> arr3[i];
    x1 = arr1[0];
    cout << x1 << " ";
    for (i = 1; i < n; i++) {
      if (arr1[i] != x1) {
        x1 = arr1[i];
      } else if (arr2[i] != x1) {
        x1 = arr2[i];
      } else if (arr3[i] != x1) {
        x1 = arr3[i];
      }
      cout << x1 << " ";
    }
    cout << endl;
  }
}
