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;
  vector<int> v1;
  vector<int> v2;
  set<int> s;
  for (int k = 0; k < t; ++k) {
    int n;
    bool f = true;
    cin >> n;
    v2.resize(2 * n);
    for (int i = 0; i < 2 * n; ++i) {
      s.insert(i + 1);
    }
    for (int i = 0; i < n; ++i) {
      cin >> v2[2 * i];
      if (s.find(v2[2 * i]) == s.end()) {
        f = false;
        break;
      } else {
        s.erase(v2[2 * i]);
      }
    }
    if (!f) {
      cout << -1 << endl;
      continue;
    } else {
      for (int i = 0; i < n; ++i) {
        auto ptr = s.lower_bound(v2[2 * i]);
        if (ptr == s.end()) {
          cout << -1 << endl;
          f = false;
          break;
        }
        v2[2 * i + 1] = *ptr;
        s.erase(ptr);
      }
      if (f) {
        for (auto x : v2) {
          cout << x << " ";
        }
        cout << endl;
      }
    }
  }
}