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;
const int MAX = 2e5 + 10;
const int MOD = 1000000007;
const double pi = acos(-1);
const int dx[] = {0, 1, -1, 0};
const int dy[] = {-1, 0, 0, 1};
const int dxc[] = {0, 2, -2, 0, 2, 2, -2, -2};
const int dyc[] = {-2, 0, 0, 2, -2, 2, -2, 2};
long long min(long long a, long long b) {
  if (a > b) return b;
  return a;
}
long long max(long long a, long long b) {
  if (a > b) return a;
  return b;
}
long long gcd(long long a, long long b) {
  while (b != 0) {
    a %= b;
    swap(a, b);
  }
  return a;
}
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
bool isPrime(long long n) {
  if (n <= 1) return false;
  if (n == 2) return true;
  if (n % 2 == 0) return false;
  for (long long i = 3; i * i <= n; i += 2) {
    if (n % i == 0) return false;
  }
  return true;
}
set<long long> gen(long long n) {
  set<long long> divs;
  for (long long i = 1; i * i <= n; i++) {
    if (n % i == 0) {
      divs.insert(i);
      divs.insert(n / i);
    }
  }
  return divs;
}
map<long long, int> fact(long long n) {
  map<long long, int> first;
  for (long long p = 2; p * p <= n; p++) {
    while (n % p == 0) {
      first[p]++;
      n /= p;
    }
  }
  if (n > 1) first[n]++;
  return first;
}
long long sum(long long n) { return (n * (n + 1)) / 2; }
long long powN(long long b, long long e) {
  long long res = 1;
  for (long long i = 0; i < e; i += 1) res *= b;
  return res;
}
long long binaryExp(long long a, long long b, long long c) {
  if (b == 0) return 1;
  if (b % 2 == 0) {
    long long temp = binaryExp(a, b / 2, c);
    return (temp * temp) % c;
  } else {
    long long temp = binaryExp(a, b - 1, c);
    return (temp * a) % c;
  }
}
bool isPal(string& cad) {
  int p1 = (int)(cad).size() - 1;
  for (long long i = 0; i < (int)(cad).size() / 2; i += 1) {
    if (cad[i] != cad[p1]) return false;
    p1--;
  }
  return true;
}
double dis(double x1, double y1, double x2, double y2) {
  double ff = pow(x2 - x1, 2) + pow(y2 - y1, 2);
  return sqrt(ff);
}
string toBin(int n) {
  string res = "";
  while (n > 0) {
    res += to_string(n % 2);
    n /= 2;
  }
  for (long long i = (int)(res).size(); i < 8; i += 1) res += "0";
  reverse((res).begin(), (res).end());
  return res;
}
long long sumDig(long long n) {
  long long res = 0;
  while (n > 0) {
    res += n % 10;
    n /= 10;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    int mat[3][n];
    for (long long i = 0; i < 3; i += 1) {
      for (long long j = 0; j < n; j += 1) cin >> mat[i][j];
    }
    vector<int> res;
    res.push_back(mat[0][0]);
    int ant = mat[0][0];
    int aux = ant;
    for (long long i = 1; i < n - 1; i += 1) {
      int mm = 0;
      for (long long j = 0; j < 3; j += 1) {
        int act = mat[j][i];
        if (act != ant) {
          mm = act;
          break;
        }
      }
      res.push_back(mm);
      ant = mm;
    }
    for (long long j = 0; j < 3; j += 1) {
      int act = mat[j][n - 1];
      if (act != aux and act != ant) {
        res.push_back(act);
        break;
      }
    }
    for (long long x = 0; x < (int)(res).size(); x += 1) cout << res[x] << " ";
    cout << "\n";
  }
  return 0;
}
