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 int64_t inf = 2e9;
int64_t inp() {
  int64_t x;
  cin >> x;
  return x;
}
string inp_str() {
  string x;
  cin >> x;
  return x;
}
int64_t sqr(int64_t n) { return n * n; }
int64_t power(int64_t a, int64_t b, int64_t mod = inf) {
  int64_t res = 1;
  a %= mod;
  for (int64_t i = b; i > 0; i /= 2) {
    if (i % 2) res = (res * a) % mod;
    a = (a * a) % mod;
  }
  return res;
}
int64_t gcd(int64_t a, int64_t b) {
  while (b > 0) {
    int64_t tmp = a % b;
    a = b;
    b = tmp;
  }
  return a;
}
bool is_pal(string s) { return equal(begin(s), end(s), s.rbegin()); }
void fast_io() {
  ios::sync_with_stdio(0);
  cin.tie(0);
}
void file_io(char *in, char *out) {
  freopen(in, "r", stdin);
  freopen(out, "w", stdout);
}
void print_bool(bool a) {
  if (a)
    cout << "YES\n";
  else
    cout << "NO\n";
}
int64_t n;
vector<int64_t> a;
void Input() {
  cin >> n;
  a = vector<int64_t>();
  a.resize(2 * n + 1);
  generate(1 + begin(a), end(a), inp);
}
void Solve() {
  vector<int64_t> visited(n + 1, 0);
  vector<int64_t> ans;
  for (int64_t i = (1); i <= (2 * n); ++i)
    if (!visited[a[i]]) {
      visited[a[i]] = true;
      ans.push_back(a[i]);
    }
  for (auto v : ans) cout << v << " ";
  cout << "\n";
}
int main() {
  fast_io();
  int64_t t = inp();
  while (t--) Input(), Solve();
  return 0;
}
