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() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  auto solve = []() {
    int n;
    cin >> n;
    vector<pair<int, int>> a(n);
    for (int i = 0; i < n; i++) {
      cin >> a[i].first;
      a[i].first--;
    }
    vector<int> ans(n + n);
    vector<int> used(n + n);
    for (int i = 0; i < n; i++) {
      ans[i * 2] = a[i].first;
      a[i].second = i * 2;
      used[a[i].first] = 1;
    }
    for (int i = 1; i < n + n; i += 2) {
      int x = ans[i - 1];
      int j;
      for (j = x + 1; j < n + n && used[j]; j++)
        ;
      if (j == n + n) {
        cout << -1 << '\n';
        return 0;
      }
      used[j] = 1;
      ans[i] = j;
    }
    for (int i = 0; i < n + n; i++) {
      if (i > 0) cout << " ";
      cout << ans[i] + 1;
    }
    cout << '\n';
  };
  while (t--) solve();
  return 0;
}