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 sortinrev(const pair<int, int> &a, const pair<int, int> &b) {
  return (a.first > b.first);
}
const int MAXN = 1000005;
long long int fac[MAXN];
const int mod = 1e9 + 7;
long long int expo(long long int a, long long int b) {
  if (b == 0)
    return 1;
  else if (b % 2 == 0) {
    long long int y = expo(a, b / 2);
    return (y * y) % mod;
  } else
    return (a * expo(a, b - 1)) % mod;
}
long long int modinv(long long int n) { return expo(n, mod - 2); }
long long int C(int n, int r) {
  long long int ans =
      (fac[n] * ((modinv(fac[r]) * modinv(fac[n - r])) % mod)) % mod;
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t, n;
  cin >> t;
  while (t--) {
    cin >> n;
    int a[2 * n];
    int h[n + 1];
    memset(h, 0, sizeof(h));
    vector<int> p;
    for (int i = 0; i < 2 * n; i++) {
      cin >> a[i];
      if (h[a[i]] == 0) {
        h[a[i]] = i + 1;
        p.push_back(i);
      }
    }
    for (auto u : p) cout << a[u] << " ";
    cout << "\n";
  }
  return 0;
}