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);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int t;
  cin >> t;
  int vis[205], p = 0;
  memset(vis, 0, sizeof(vis));
  while (t--) {
    ++p;
    int n, b[205];
    cin >> n;
    for (int i = 0; i < n; ++i) {
      cin >> b[i];
      vis[b[i]] = p;
    }
    int index;
    bool f = false;
    vector<int> ans;
    for (int i = 0; i < n; ++i) {
      index = -1;
      for (int j = b[i] + 1; j <= 2 * n; ++j) {
        if (vis[j] != p) {
          index = j;
          break;
        }
      }
      if (index == -1) {
        f = true;
        break;
      }
      vis[index] = p;
      ans.push_back(b[i]);
      ans.push_back(index);
    }
    if (!f) {
      for (int i = 0; i < ans.size(); ++i) cout << ans[i] << " ";
    } else
      cout << -1;
    cout << endl;
  }
  return 0;
}