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>
void print(T t) {
  cout << t << endl;
}
long long n = 0, t = 0;
long long small = -1e18, big = 1e18;
const long long mod = 1000000007;
map<long long, long long> occ;
const long long N = 2e5;
map<pair<long long, long long>, long long> occP;
long long factorial(int n) {
  long long ans = 1;
  for (int i = 1; i <= n; ++i) {
    ans *= i;
  }
  return ans;
}
long long findGCD(long long a, long long b) {
  long long ans = 1;
  for (int i = 1; i <= min(a, b); ++i) {
    if (a % i == 0 && b % i == 0) {
      ans = i;
    }
  }
  return ans;
}
bool sortVec(const vector<long long>& v1, const vector<long long>& v2) {
  return v1[0] < v2[0];
}
bool isprime(long long p) {
  for (int i = 2; i < pow(p, 1.5); ++i) {
    if (p % i == 0 && p != i) {
      return false;
    }
  }
  return true;
}
void print2dVec(vector<vector<long long>> v) {
  for (int i = (0); i < (v.size()); ++i) {
    cout << i << " : ";
    for (int j = (0); j < (v[i].size()); ++j) {
      cout << v[i][j] << " ";
    }
    print("");
  }
}
bool check(long long val) { return false; }
long long binsearch(long long low, long long high) {
  while (low < high) {
    long long mid = (low + high) / 2;
    if (check(mid)) {
      high = mid;
    } else {
      low = mid + 1;
    }
  }
  return low;
}
long long findLCM(long long a, long long b) {
  for (long long i = min(a, b); i <= a * b; ++i) {
    if (i % a == 0 && i % b == 0) {
      return i;
    }
  }
  return -1;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << fixed;
  cout << setprecision(10);
  ;
  cin >> t;
  while (t--) {
    cin >> n;
    vector<long long> a(n), b(n), c(n);
    for (long long it = 0; it < n; it++) cin >> a[it];
    ;
    for (long long it = 0; it < n; it++) cin >> b[it];
    ;
    for (long long it = 0; it < n; it++) cin >> c[it];
    ;
    long long taken = 1;
    cout << a[0] << " ";
    long long prev = a[0];
    while (taken < n - 1) {
      if (a[taken] != prev) {
        cout << a[taken] << " ";
        prev = a[taken];
      } else if (b[taken] != prev) {
        cout << b[taken] << " ";
        prev = b[taken];
      } else {
        cout << c[taken] << " ";
        prev = c[taken];
      }
      taken += 1;
    }
    if (c[taken] != prev && c[taken] != a[0]) {
      cout << c[taken];
    } else {
      cout << b[taken];
    }
    print("");
  }
  return 0;
}
