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;
signed main() {
  long long tt;
  cin >> tt;
  while (tt--) {
    long long n;
    cin >> n;
    vector<long long> a(2 * n), b(n);
    set<long long> help;
    for (long long i = 0; i < n; ++i) {
      cin >> b[i];
      help.insert(2 * i);
      help.insert(2 * i + 1);
    }
    vector<pair<long long, long long> > all(n);
    for (long long i = 0; i < n; ++i) {
      a[2 * i] = b[i] - 1;
      help.erase(b[i] - 1);
      all[i] = {b[i] - 1, i};
    }
    sort(all.begin(), all.end());
    long long i = 0;
    bool ans = true;
    for (auto u : help) {
      if (a[all[i].second * 2] > u) {
        cout << -1;
        ans = false;
        break;
      }
      a[all[i].second * 2 + 1] = u;
      ++i;
    }
    if (ans) {
      for (auto u : help) {
        all.push_back({u, -1});
      }
      sort(all.begin(), all.end());
      set<long long> now;
      for (i = 0; i < all.size(); ++i) {
        if (all[i].second != -1)
          now.insert(all[i].second);
        else {
          auto k = *now.begin();
          a[2 * k + 1] = all[i].first;
          now.erase(k);
        }
      }
      for (i = 0; i < 2 * n; ++i) cout << a[i] + 1 << " ";
    }
    cout << "\n";
  }
}