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;
int main() {
  int t;
  cin >> t;
  while (t > 0) {
    t--;
    int n;
    cin >> n;
    int giv[2 * n], give[2 * n];
    for (int i = 0; i < 2 * n; i++) {
      cin >> giv[i];
      give[i] = giv[i];
    }
    vector<int> a;
    a.push_back(giv[0]);
    give[0] = 0;
    for (int i = 0; i < 2 * n; i++) {
      if (giv[i] != 0) {
        if (give[i] != 0) a.push_back(giv[i]);
        for (int j = i + 1; j < 2 * n; j++) {
          if (giv[i] != giv[j] && give[j] != 0) {
            a.push_back(giv[j]);
            give[j] = 0;
          } else if (giv[i] == giv[j]) {
            giv[j] = 0;
            give[j] = 0;
            break;
          }
        }
      }
    }
    for (int i = 0; i < n; i++) cout << a[i] << " ";
    cout << endl;
  }
}