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 arr[1005];
int counter[1005];
int main() {
  int tc;
  cin >> tc;
  while (tc--) {
    memset(arr, 0, 1005);
    memset(counter, 0, 1005);
    int flag = 0;
    vector<int> v;
    int n;
    cin >> n;
    for (int i = 0; i < n; i++) {
      cin >> arr[i];
      counter[arr[i]]++;
    }
    int lim = 2 * n;
    for (int i = 0; i < n; i++) {
      v.push_back(arr[i]);
      for (int j = arr[i] + 1;; j++) {
        if (j > lim) {
          flag = 1;
          break;
        }
        if (!counter[j]) {
          counter[j]++;
          v.push_back(j);
          break;
        }
      }
    }
    if (flag == 1)
      cout << "-1" << endl;
    else {
      for (auto it = v.begin(); it != v.end(); ++it) cout << *it << " ";
      cout << endl;
    }
  }
}