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(false);
  cin.tie(0);
  cout.tie(0);
  int T;
  cin >> T;
  while (T--) {
    int n;
    cin >> n;
    int aa[n + 1], cc[n + 1], bb[n * 2 + 2], dd[2 * n + 2];
    for (int i = 0; i < 2 * n + 2; i++) bb[i] = 1;
    for (int i = 0; i < n; i++) {
      cin >> aa[i];
      cc[i] = aa[i];
      bb[aa[i]] = 0;
    }
    sort(cc, cc + n);
    bool f = true;
    if (cc[0] != 1 && cc[n - 1] == 2 * n) f = false;
    int j;
    for (int i = 0; i < n; i++) {
      j = aa[i];
      while (!bb[j] && j <= 2 * n) j++;
      if (j > 2 * n) {
        f = false;
        break;
      }
      dd[i * 2] = aa[i];
      dd[i * 2 + 1] = j;
      bb[j] = 0;
    }
    if (!f)
      cout << "-1" << endl;
    else {
      for (int i = 0; i < 2 * n; i++) cout << dd[i] << " ";
      cout << endl;
    }
  }
  return 0;
  ;
}