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;
template <typename T>
T BigMod(T b, T p, T m) {
  if (p == 0) return 1;
  if (p % 2 == 0) {
    T s = BigMod(b, p / 2, m);
    return ((s % m) * (s % m)) % m;
  }
  return ((b % m) * (BigMod(b, p - 1, m) % m)) % m;
}
template <typename T>
T POW(T B, T P) {
  if (P == 0) return 1;
  if (P & 1)
    return B * POW(B, P - 1);
  else
    return POW(B, P / 2) * POW(B, P / 2);
}
template <typename T>
T gcd(T a, T b) {
  if (a < 0) return gcd(-a, b);
  if (b < 0) return gcd(a, -b);
  return (b == 0) ? a : gcd(b, a % b);
}
template <typename T>
T lcm(T a, T b) {
  if (a < 0) return lcm(-a, b);
  if (b < 0) return lcm(a, -b);
  return a * (b / gcd(a, b));
}
string to_bin(long long int num) {
  string bin = "";
  while (num) {
    if (num & 1)
      bin = "1" + bin;
    else
      bin = "0" + bin;
    num = num >> 1;
  }
  return bin;
}
int b2d(long long n) {
  int decimalNumber = 0, i = 0, remainder;
  while (n != 0) {
    remainder = n % 10;
    n /= 10;
    decimalNumber += remainder * pow(2, i);
    ++i;
  }
  return decimalNumber;
}
int toInt(string s) {
  int sm;
  stringstream ss(s);
  ss >> sm;
  return sm;
}
long long toLlint(string s) {
  long long sm;
  stringstream ss(s);
  ss >> sm;
  return sm;
}
bool isVowel(char ch) {
  ch = toupper(ch);
  if (ch == 'A' || ch == 'U' || ch == 'I' || ch == 'O' || ch == 'E')
    return true;
  return false;
}
template <class T>
string NTS(T Number) {
  stringstream ss;
  ss << Number;
  return ss.str();
}
const int N = 1000000;
bitset<N + 1> bs;
vector<int> prime;
int prifix[N + 1];
void sieve() {
  bs.set();
  bs[0] = bs[1] = 0;
  for (int i = 2; i * i <= N; ++i) {
    if (bs[i]) {
      prime.push_back(i);
      for (int j = i * i; j <= N; j += i) bs[j] = 0;
    }
  }
  prifix[0] = 0;
  prifix[1] = 0;
  for (int i = 2; i <= N; i++) {
    prifix[i] += prifix[i - 1];
    if (bs[i]) prifix[i]++;
  }
}
void prime_divisor(long long n, long long q) {
  if (n % q)
    cout << n << "\n";
  else {
    long long nn = n;
    for (long long i = 1; i <= sqrt(n); i++) {
      if (n % i == 0) {
        {
          if ((n / i) % q != 0) {
            cout << n / i << "\n";
            break;
          }
        }
      }
    }
  }
}
unsigned long long fact(unsigned long long n) {
  if (n <= 1) return 1;
  return fact(n - 1) * n;
}
long long sumDigits(long long n) {
  long long sum = 0;
  while (n != 0) {
    sum += n % 10;
    n /= 10;
  }
  return sum;
}
long long int toDec(string bin) {
  long long int num = 0;
  for (int i = 0; i < bin.length(); i++) {
    if (bin.at(i) == '1') num += 1 << (bin.length() - i - 1);
  }
  return num;
}
vector<long long> ans;
int ho;
bool isprime(int w) {
  if (w == 0) return false;
  if (w == 1) return false;
  int sqrtw = int(sqrt((double)w));
  for (int q = 2; q <= sqrtw; q++) {
    if (w % q == 0) {
      ho = q;
      return false;
    }
  }
  return true;
}
int main() {
  {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
  }
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    long long ar1[n], ar2[n], ar3[n], ans[n];
    for (long long i = 0; i < n; i++) {
      cin >> ar1[i];
    }
    for (long long i = 0; i < n; i++) {
      cin >> ar2[i];
    }
    for (long long i = 0; i < n; i++) {
      cin >> ar3[i];
    }
    ans[0] = ar1[0];
    for (long long i = 1; i < n; i++) {
      if (ans[i - 1] == ar1[i])
        ans[i] = ar2[i];
      else
        ans[i] = ar1[i];
    }
    for (long long i = 0; i < n; i++) {
      if (ans[n - 1] != ans[n - 2] and ans[n - 1] != ans[0])
        break;
      else
        ans[n - 1] = ar3[i];
    }
    for (int a : ans) cout << a << " ";
    cout << endl;
  }
}
