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() {
  int n, i, test;
  cin >> test;
  int tc = 0;
  while (tc < test) {
    cin >> n;
    int arr[n], f[2 * n];
    for (i = 0; i < n; cin >> arr[i], i++)
      ;
    vector<int> mark;
    mark.assign(2 * n + 1, 0);
    for (i = 0; i < n; i++) {
      f[2 * i] = arr[i];
      f[(2 * i) + 1] = arr[i];
      mark[arr[i]] = 1;
    }
    for (i = 0; i < n; i++) {
      int num = f[2 * i + 1];
      while (num <= 2 * n) {
        if (mark[num] == 0) {
          f[(2 * i) + 1] = num;
          mark[num] = 1;
          break;
        }
        num++;
      }
    }
    int status = 1;
    for (i = 1; i <= 2 * n; i++) {
      if (mark[i] == 0) {
        status = 0;
      }
    }
    if (status == 1)
      for (i = 0; i < 2 * n; cout << f[i] << " " << endl, i++)
        ;
    else
      cout << "-1" << endl;
    tc++;
  }
  return 0;
}