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_base::sync_with_stdio(0);
  istream::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int q;
  cin >> q;
  for (; q > 0; q--) {
    int n;
    cin >> n;
    int b[n];
    bool mk[n + n + 10];
    for (int i = 0; i <= n + n; i++) mk[i] = 0;
    for (int i = 0; i < n; i++) {
      cin >> b[i];
      mk[b[i]] = 1;
    }
    int a[n + n];
    bool gd = 1;
    for (int i = 0; i < n; i++) {
      int nm = -1;
      for (int j = b[i]; j <= n + n; j++)
        if (!mk[j]) {
          nm = j;
          break;
        }
      if (nm == -1) {
        gd = 0;
        cout << -1 << endl;
        break;
      }
      mk[nm] = 1;
      a[(i + 1) * 2 - 1] = nm;
      a[(i + 1) * 2 - 2] = b[i];
    }
    if (!gd) continue;
    for (int i = 0; i < n + n; i++) cout << a[i] << " ";
    cout << endl;
  }
}