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 i, j, k, n, t, c;
  cin >> t;
  for (; t > 0; t--) {
    c = 1;
    cin >> n;
    int a[2 * n];
    bool x[2 * n + 1];
    memset(x, 0, sizeof(x));
    j = 1;
    for (i = 0; i < n; i++) {
      cin >> k;
      a[2 * i] = k;
      x[k] = 1;
    }
    for (i = 0; i < n; i++) {
      j = a[2 * i];
      j++;
      for (; x[j] == 1 && j <= 2 * n; j++)
        ;
      if (j <= 2 * n) {
        a[2 * i + 1] = j;
        x[j] = 1;
      } else
        break;
    }
    if (i == n)
      for (i = 0; i < n; i++) cout << a[2 * i] << " " << a[2 * i + 1] << " ";
    else
      cout << -1;
    cout << endl;
  }
}