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;
const long double pi = 3.14159265358979323;
long long modu = 1000000000 + 7;
long long mod = 998244353;
const long double pii = acos(-1.0);
const long long INF = 1e18;
const long long inf = 1e9;
long long power(long long x, long long y) {
  long long res = 1;
  x = x;
  while (y > 0) {
    if (y & 1) res = (res * x);
    y = y >> 1;
    x = (x * x);
  }
  return res;
}
long long powe(long long x, long long y) {
  x = x % mod, y = y % (mod - 1);
  long long ans = 1;
  while (y > 0) {
    if (y & 1) {
      ans = (1ll * x * ans) % mod;
    }
    y >>= 1;
    x = (1ll * x * x) % mod;
  }
  return ans;
}
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long lcm(long long a, long long b) { return (a / gcd(a, b) * b); }
bool isPrime(long long n) {
  if (n < 2) return false;
  for (long long i = 2; i * i <= n; i++)
    if (n % i == 0) return false;
  return true;
}
long long ncr(long long n, long long r) {
  long long res = 1;
  if (r > n) return 0;
  if (r > n - r) r = n - r;
  for (long long i = 0; i < r; i++) {
    res *= (n - i);
    res /= (i + 1);
  }
  return res;
}
long long add(long long a, long long b) { return ((a % mod + b % mod) % mod); }
long long sub(long long a, long long b) {
  return ((a % mod - b % mod + mod) % mod);
}
long long mul(long long a, long long b) {
  return (((a % mod) * (b % mod)) % mod);
}
long long divi(long long a, long long b) {
  return (mul(a, powe(b, mod - 2)) % mod);
}
void fun() {}
const long long N = 1e3 + 7;
long long a[N][N];
void pre() {}
bool ok(char ch) {
  return (ch == '(' or ch == ')' or ch == '{' or ch == '}' or ch == '[' or
          ch == ']');
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  fun();
  pre();
  long long tttt = 1;
  cin >> tttt;
  while (tttt--) {
    long long n;
    cin >> n;
    long long a[2 * n];
    set<long long> s;
    for (long long i = 0; i < 2 * n; i++) {
      long long x;
      cin >> x;
      if (s.find(x) == s.end()) cout << x << ' ', s.insert(x);
    }
    cout << "\n";
  }
  cerr << "time elapsed : " << 1.0 * clock() / CLOCKS_PER_SEC << " sec \n";
  return 0;
}
