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() {
  ios_base::sync_with_stdio(false);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    int inp[n];
    set<int> s;
    for (int i = 0; i < n; i++) {
      cin >> inp[i];
      s.insert(inp[i]);
    }
    vector<pair<int, int> > ans;
    int act;
    for (int i = 0; i < n; i++) {
      act = inp[i] + 1;
      while (act <= 2 * n) {
        if (s.count(act) == 0) {
          ans.push_back(pair<int, int>(inp[i], act));
          s.insert(act);
          break;
        } else
          act++;
      }
      if (act > 2 * n) break;
    }
    if (act > 2 * n)
      cout << "-1\n";
    else {
      for (int i = 0; i < n; i++) {
        if (i == n - 1)
          cout << ans[i].first << " " << ans[i].second << "\n";
        else
          cout << ans[i].first << " " << ans[i].second << " ";
      }
    }
  }
  return 0;
}