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;
const int MOD = 1e9 + 7;
const int N = 2e5 + 5;
template <typename T>
inline istream& operator>>(istream& is, vector<T>& v) {
  for (int ii = 0; ii < (int)v.size(); ++ii) {
    is >> v[ii];
  }
  return is;
}
template <typename T>
inline ostream& operator<<(ostream& os, vector<T>& v) {
  for (int ii = 0; ii < (int)v.size(); ++ii) {
    os << v[ii] << " ";
  }
  return os;
}
template <typename T1, typename T2>
inline istream& operator>>(istream& is, pair<T1, T2>& p) {
  is >> p.first >> p.second;
  return is;
}
template <typename T>
inline ostream& operator<<(ostream& os, vector<vector<T>>& mat) {
  for (int ii = 0; ii < (int)mat.size(); ++ii) {
    for (int jj = 0; jj < (int)mat[0].size(); ++jj) {
      os << mat[ii][jj] << " ";
    }
    os << '\n';
  }
  return os;
}
template <typename it>
void compress(it start, it end) {
  typedef typename remove_reference<decltype(*start)>::type T;
  map<T, vector<it>> m;
  for (it ii = start; ii != end; ii++) {
    m[*ii].push_back(ii);
  }
  int t = 0;
  for (auto& [x, v] : m) {
    for (auto& i : v) {
      *i = t;
    }
    ++t;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<int> a(2 * n);
    cin >> a;
    vector<int> b;
    vector<bool> was(n + 1, false);
    for (int ii = 0; ii < (int)a.size(); ++ii) {
      if (!was[a[ii]]) {
        was[a[ii]] = true;
        b.push_back(a[ii]);
      }
    }
    cout << b << '\n';
  }
  return 0;
}
