Problem: You are given a sequence b_1, b_2, …, b_n. Find the lexicographically minimal permutation a_1, a_2, …, a_{2n} such that b_i = min(a_{2i-1}, a_{2i}), or determine that it is impossible.

Input

Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).

The first line of each test case consists of one integer n — the number of elements in the sequence b (1 ≤ n ≤ 100).

The second line of each test case consists of n different integers b_1, …, b_n — elements of the sequence b (1 ≤ b_i ≤ 2n).

It is guaranteed that the sum of n by all test cases doesn't exceed 100.

Output

For each test case, if there is no appropriate permutation, print one number -1.

Otherwise, print 2n integers a_1, …, a_{2n} — required lexicographically minimal permutation of numbers from 1 to 2n.

Example

Input


5
1
1
2
4 1
3
4 1 3
4
2 3 4 5
5
1 5 7 2 8


Output


1 2 
-1
4 5 1 2 3 6 
-1
1 3 5 6 7 9 2 4 8 10 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(0);
  std::cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<int> a(n), b(2 * n);
    multiset<int> m;
    for (size_t i = 0; i < 2 * n; ++i) m.insert(i + 1);
    for (auto& i : a) {
      cin >> i;
      m.erase(i);
    }
    bool flg = true;
    for (size_t i = 0; i < n; ++i) {
      b[2 * i] = a[i];
      auto it = upper_bound(m.begin(), m.end(), a[i]);
      if (it == m.end()) {
        flg = false;
        cout << -1 << endl;
        break;
      }
      b[2 * i + 1] = *it;
      m.erase(*it);
    }
    if (!flg) continue;
    for (auto& i : b) cout << i << " ";
    cout << endl;
  }
  return 0;
}