Problem: 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 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
bool isprime(long long n) {
  if (n <= 1) return false;
  long long g = sqrt(n);
  for (long long i = 2; i <= g; i++)
    if (n % i == 0) return false;
  return true;
}
long long power(long long base, long long exp) {
  long long MOD = LLONG_MAX;
  if (exp == 1)
    return base;
  else {
    if (exp % 2 == 0) {
      long long base1 = (pow(power(base, exp / 2), 2));
      return base1 % MOD;
    } else {
      long long ans = (base * pow(power(base, (exp - 1) / 2), 2));
      return ans % MOD;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long temp, temp1, temp2;
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    set<int> s;
    for (int i = 1; i <= n; i++) {
      s.insert(i);
    }
    auto itr = s.begin();
    vector<int> v;
    for (int i = 0; i < 2 * n; i++) {
      cin >> temp;
      itr = find(s.begin(), s.end(), temp);
      if (itr != s.end()) {
        s.erase(itr);
        v.push_back(temp);
      }
    }
    for (long long i = 0; i < n; i++) {
      cout << v[i] << " ";
    }
    cout << endl;
  }
  return 0;
}