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;
template <class T>
inline T bigmod(T b, T p, T M) {
  T ret = 1;
  for (; p > 0; p >>= 1) {
    if (p & 1) ret = (ret * b) % M;
    b = (b * b) % M;
  }
  return (T)ret;
}
template <class T>
inline T modinverse(T a, T M) {
  return bigmod(a, M - 2, M);
}
template <class T>
inline T gcd(T a, T b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
template <class T>
inline T lcm(T a, T b) {
  return a * b / gcd<T>(a, b);
}
template <class T>
inline T power(T b, T p) {
  long long int x = 1;
  while (p--) x *= b;
  return x;
}
const int mod = 1000000007;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  ;
  int tc;
  cin >> tc;
  while (tc--) {
    int n;
    cin >> n;
    unordered_map<int, int> vis;
    vector<int> vec;
    for (int i = 1; i <= 2 * n; i++) {
      int a;
      cin >> a;
      if (vis[a] == 0) vec.push_back(a);
      vis[a] = 1;
    }
    for (int i = 0; i < vec.size(); i++) cout << vec[i] << " ";
    cout << endl;
    ;
  }
  return 0;
}
