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;
long long poww(long long x, long long y, long long m) {
  x = x % m;
  if (y == 0) {
    return 1;
  }
  if (y % 2 == 0) {
    return poww(x * x, y / 2, m);
  } else {
    return (x * poww(x * x, y / 2, m)) % m;
  }
}
const int mod = 1e9 + 7;
const int maxn = 2e5 + 10;
int t;
int n, a[maxn];
void solve() {
  vector<int> a1;
  vector<int> a2;
  int i, j;
  int first = 0, second = 0;
  a1.push_back(a[0]);
  for ((i) = 1; (i) < (2 * n); i = i + 1) {
    if (a[i] == a1[first] && first < a1.size()) {
      a2.push_back(a[i]);
      first++;
    } else {
      a1.push_back(a[i]);
    }
  }
  for ((i) = 0; (i) < (n); i = i + 1) {
    cout << a1[i] << " ";
  }
  cout << "\n";
  ;
}
int main() {
  int i, j;
  cin >> t;
  while (t--) {
    cin >> n;
    for ((i) = 0; (i) < (2 * n); i = i + 1) {
      cin >> a[i];
    }
    solve();
  }
  return 0;
}