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::sync_with_stdio(false);
  cin.tie(0);
  long long int n;
  int t;
  cin >> t;
  while (t--) {
    cin >> n;
    vector<long long int> v(n), vv;
    for (long long int i = 0; i <= n - 1; i++) cin >> v[i];
    bool arr[210];
    memset(arr, 0, sizeof(arr));
    bool ok = 0, ans = 1;
    for (long long int i = 0; i <= n - 1; i++) arr[v[i]] = 1;
    for (long long int i = 0; i < n; i++) {
      ok = 0;
      vv.push_back(v[i]);
      for (long long int j = v[i] + 1; j <= 2 * n; j++) {
        if (arr[j] == 0) {
          vv.push_back(j);
          arr[j] = 1;
          ok = 1;
          break;
        }
      }
      if (!ok) {
        ans = 0;
        break;
      }
    }
    if (ans) {
      for (long long int i = 0; i <= n + n - 1; i++) cout << vv[i] << ' ';
      cout << endl;
    } else
      cout << -1 << endl;
  }
  return 0;
}