A permutation of length n is a sequence of integers from 1 to n of length n containing each number exactly once. For example, [1], [4, 3, 5, 1, 2], [3, 2, 1] are permutations, and [1, 1], [0, 1], [2, 2, 1, 4] are not.

There was a permutation p[1 ... n]. It was merged with itself. In other words, let's take two instances of p and insert elements of the second p into the first maintaining relative order of elements. The result is a sequence of the length 2n.

For example, if p=[3, 1, 2] some possible results are: [3, 1, 2, 3, 1, 2], [3, 3, 1, 1, 2, 2], [3, 1, 3, 1, 2, 2]. The following sequences are not possible results of a merging: [1, 3, 2, 1, 2, 3], [3, 1, 2, 3, 2, 1], [3, 3, 1, 2, 2, 1].

For example, if p=[2, 1] the possible results are: [2, 2, 1, 1], [2, 1, 2, 1]. The following sequences are not possible results of a merging: [1, 1, 2, 2], [2, 1, 1, 2], [1, 2, 2, 1].

Your task is to restore the permutation p by the given resulting sequence a. It is guaranteed that the answer exists and is unique.

You have to answer t independent test cases.

Input

The first line of the input contains one integer t (1 ≤ t ≤ 400) — the number of test cases. Then t test cases follow.

The first line of the test case contains one integer n (1 ≤ n ≤ 50) — the length of permutation. The second line of the test case contains 2n integers a_1, a_2, ..., a_{2n} (1 ≤ a_i ≤ n), where a_i is the i-th element of a. It is guaranteed that the array a represents the result of merging of some permutation p with the same permutation p.

Output

For each test case, print the answer: n integers p_1, p_2, ..., p_n (1 ≤ p_i ≤ n), representing the initial permutation. It is guaranteed that the answer exists and is unique.

Example

Input


5
2
1 1 2 2
4
1 3 1 4 3 4 2 2
5
1 2 1 2 3 4 3 5 4 5
3
1 2 3 1 2 3
4
2 3 2 4 1 3 4 1


Output


1 2 
1 3 4 2 
1 2 3 4 5 
1 2 3 
2 3 4 1 
#include <bits/stdc++.h>
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const long long MOD = 1000000007;
const long long inf = 1e18;
const long long MOD_1 = 998244353;
inline long long gcd(long long a, long long b) {
  if (a == 0)
    return b;
  else
    return gcd(b % a, a);
}
inline void take_input(long long A[], long long n) {
  for (long long i = 0; i < n; i++) {
    cin >> A[i];
  }
}
inline void show(long long *A, long long n) {
  for (long long i = 0; i < n; i++) {
    cout << A[i] << " ";
  }
  cout << '\n';
}
inline long long countDivisibles(long long a, long long b, long long c) {
  if (a % c == 0) {
    return (b / c - a / c + 1);
  }
  return (b / c - a / c);
}
inline bool sortbysec(const pair<long long, long long> &a,
                      const pair<long long, long long> &b) {
  return (a.second < b.second);
}
inline bool sortbysecdesc(const pair<long long, long long> &a,
                          const pair<long long, long long> &b) {
  return (a.second > b.second);
}
inline void swap(long long &a, long long &b) {
  long long temp = a;
  a = b;
  b = temp;
}
inline void initialise(long long A[], long long n, long long k = 0) {
  for (long long i = 0; i < n; i++) {
    A[i] = k;
  }
}
inline long long modulo(long long a, long long b) { return (b + (a % b)) % b; }
inline long long getascii(char ch) { return (long long)ch; }
inline long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
template <typename T>
void show_vect(vector<T> v) {
  for (long long i = 0; i < v.size(); i++) {
    cout << v[i] << " ";
  }
  cout << '\n';
}
template <typename T>
void vect_input(vector<T> &v, T n) {
  for (long long i = 0; i < n; i++) {
    cin >> v[i];
  }
}
template <typename T>
void show_set(set<T> s) {
  for (auto x : s) {
    cout << x << " ";
  }
  cout << '\n';
}
const long long rownbr[8] = {-1, -1, -1, 0, 0, 1, 1, 1};
const long long colnbr[8] = {-1, 0, 1, -1, 1, -1, 0, 1};
void p_g_t() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
int32_t main() {
  p_g_t();
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    vector<long long> v(2 * n);
    vect_input<long long>(v, 2 * n);
    vector<long long> ans;
    for (long long i = 0; i < 2 * n; i++) {
      if (find(ans.begin(), ans.end(), v[i]) == ans.end()) ans.push_back(v[i]);
    }
    show_vect<long long>(ans);
  }
  return 0;
}
