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 b[100];
int a[201];
int main(void) {
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    set<int> left;
    for (int i = 1; i <= (2 * n); i++) {
      left.insert(i);
    }
    for (int i = 0; i < n; i++) {
      cin >> b[i];
      left.erase(b[i]);
    }
    int can = 1;
    for (int i = 0; i < n; i++) {
      a[2 * i] = b[i];
      auto it = left.upper_bound(b[i]);
      if (it == left.end()) {
        can = 0;
        break;
      }
      a[2 * i + 1] = *it;
      left.erase(it);
    }
    if (!can)
      cout << "-1\n";
    else {
      for (int i = 0; i < 2 * n; i++) {
        cout << a[i] << ' ';
      }
      cout << '\n';
    }
  }
  return 0;
}