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;
long long MOD = 1000000007;
bool isPrime(long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
long long power(long long x, unsigned long long y) {
  if (y == 0)
    return 1;
  else if (y % 2 == 0)
    return power(x, y / 2) * power(x, y / 2);
  else
    return x * power(x, y / 2) * power(x, y / 2);
}
vector<long long> decToBinary(long long n) {
  vector<long long> v;
  for (long long i = 31; i >= 0; i--) {
    long long k = n >> i;
    if (k & 1)
      v.push_back(1);
    else
      v.push_back(0);
  }
  return v;
}
long long euclid(long long a, long long b) {
  if (b == 0)
    return a;
  else
    return euclid(b, a % b);
}
long long binarySearch(long long arr[], long long l, long long r, long long x) {
  if (r >= l) {
    long long mid = l + (r - l) / 2;
    if (arr[mid] == x) return mid;
    if (arr[mid] > x) return binarySearch(arr, l, mid - 1, x);
    return binarySearch(arr, mid + 1, r, x);
  }
  return -1;
}
long long maxResult(long long n, long long a, long long b, long long c) {
  long long maxVal = 0;
  for (long long i = 0; i <= n; i += a)
    for (long long j = 0; j <= n - i; j += b) {
      float z = (float)(n - (i + j)) / c;
      if (floor(z) == ceil(z)) {
        long long x = i / a;
        long long y = j / b;
        maxVal = max(maxVal, x + y + (long long)z);
      }
    }
  return maxVal;
}
int32_t main() {
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    long long a[2 * n];
    for (long long i = 0; i < 2 * n; i++) {
      cin >> a[i];
    }
    map<long long, bool> m;
    for (long long i = 0; i < n; i++) {
      m[i + 1] = true;
    }
    vector<long long> v;
    for (long long i = 0; i < 2 * n; i++) {
      if (m[a[i]] == true) {
        v.push_back(a[i]);
        m[a[i]] = false;
      }
    }
    for (long long i = 0; i < v.size(); i++) {
      cout << v[i] << " ";
    }
    cout << endl;
  }
}
