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 t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    int b[n];
    unordered_set<int> s;
    for (int i = 0; i < n; i++) {
      cin >> b[i];
      s.insert(b[i]);
    }
    int a[2 * n], k = 0;
    for (int i = 0; i < 2 * n; i++) {
      if (i % 2 == 0) {
        a[i] = b[k++];
      }
    }
    int p = 0;
    for (int i = 1; i <= 2 * n; i++) {
      if (s.find(i) == s.end()) {
        if (a[p] < i) {
          a[p + 1] = i;
          s.insert(i);
          p += 2;
          i = 1;
        }
      }
    }
    int flag = 0;
    for (int i = 0; i < 2 * n; i++) {
      if (a[i] >= 1 && a[i] <= 2 * n) {
        flag = 0;
      } else {
        flag = 1;
        break;
      }
    }
    if (flag == 0) {
      for (int i = 0; i < 2 * n; i++) {
        cout << a[i] << " ";
      }
    } else {
      cout << "-1";
    }
    cout << endl;
  }
}