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;
inline int add(int _a, int _b) {
  if (_a < 0) {
    _a += 1000000007;
  }
  if (_b < 0) {
    _b += 1000000007;
  }
  if (_a + _b >= 1000000007) {
    return _a + _b - 1000000007;
  }
  return _a + _b;
}
inline int mul(int _a, int _b) {
  if (_a < 0) {
    _a += 1000000007;
  }
  if (_b < 0) {
    _b += 1000000007;
  }
  return ((long long int)((long long int)_a * (long long int)_b)) % 1000000007;
}
const int N = 50;
int n, ara[2 * N + 10];
bool mark[N + 5];
void solve() {
  int i, j;
  memset(mark, false, sizeof(mark));
  vector<int> vec;
  for (i = 0; i < 2 * n; ++i) {
    if (!mark[ara[i]]) {
      vec.push_back(ara[i]);
      mark[ara[i]] = true;
    }
  }
  for (i = 0; i < vec.size(); i++) {
    if (i) printf(" ");
    printf("%d", vec[i]);
  }
  puts("");
}
int main() {
  int cs, ts;
  scanf("%d", &ts);
  for (cs = 0; cs < ts; cs++) {
    int i, j;
    scanf("%d", &n);
    for (i = 0; i < 2 * n; i++) scanf("%d", &ara[i]);
    solve();
  }
}