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 max(long long a, long long b) {
  if (a > b)
    return a;
  else
    return b;
}
long long min(long long a, long long b) {
  if (a < b)
    return a;
  else
    return b;
}
long long lowbit(long long x) { return x & (-x); }
long long prime(long long x) {
  if (x <= 1) return false;
  for (int i = 2; i <= int(sqrt(x)); i++) {
    if (x % i == 0) return false;
  }
  return true;
}
bool cmp(long long a, long long b) { return a > b; }
long long gcd(long long a, long long b) {
  long long r;
  while (b > 0) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
long long powmod(long long a, long long b, long long mod) {
  long long res = 1;
  a %= mod;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
inline int readint() {
  char c = getchar();
  int w = 1;
  int x = 0;
  while (!isdigit(c)) (c == '-') && (w = -w), c = getchar();
  while (isdigit(c)) x = (x << 3) + (x << 1) + (c ^ '0'), c = getchar();
  x *= w;
  return x;
}
inline long long readll() {
  char c = getchar();
  long long w = 1;
  long long x = 0;
  while (!isdigit(c)) (c == '-') && (w = -w), c = getchar();
  while (isdigit(c)) x = (x << 3) + (x << 1) + (c ^ '0'), c = getchar();
  x *= w;
  return x;
}
void fl(string name) {
  freopen((name + ".in").c_str(), "r", stdin);
  freopen((name + ".out").c_str(), "w", stdout);
}
const int dx[8] = {-1, 1, 0, 0, -1, 1, -1, 1},
          dy[8] = {0, 0, -1, 1, -1, -1, 1, 1};
int main() {
  long long q;
  cin >> q;
  while (q--) {
    long long n;
    cin >> n;
    long long a[n + 1], b[n + 1], c[n + 1], p[n + 1];
    for (long long i = 1; i <= n; i++) cin >> a[i];
    for (long long i = 1; i <= n; i++) cin >> b[i];
    for (long long i = 1; i <= n; i++) cin >> c[i];
    p[1] = a[1];
    for (long long i = 1; i <= n; i++) {
      if (a[i] != p[i - 1]) {
        p[i] = a[i];
        continue;
      }
      if (b[i] != p[i - 1]) {
        p[i] = b[i];
        continue;
      }
      if (c[i] != p[i - 1]) {
        p[i] = c[i];
        continue;
      }
    }
    long long flag = 0;
    if (a[n] != p[1] && p[n - 1] != a[n]) {
      p[n] = a[n];
      flag = 1;
    }
    if (b[n] != p[1] && p[n - 1] != b[n] && !flag) {
      p[n] = b[n];
      flag = 1;
    }
    if (c[n] != p[1] && p[n - 1] != c[n] && !flag) {
      p[n] = c[n];
    }
    for (long long i = 1; i <= n; i++) cout << p[i] << ' ';
    cout << '\n';
  }
  return 0;
}