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;
void solve() {
  int n;
  cin >> n;
  vector<int> a(2 * n);
  set<int> s;
  for (int i = int(1); i < int(2 * n + 1); ++i) {
    s.insert(i);
  }
  for (int i = 0; i < int(n); ++i) {
    cin >> a[2 * i];
    s.erase(a[2 * i]);
  }
  for (int i = 0; i < int(n); ++i) {
    set<int>::iterator it = s.upper_bound(a[2 * i]);
    if (it == s.end()) {
      cout << -1 << '\n';
      return;
    }
    a[2 * i + 1] = *it;
    s.erase(it);
  }
  for (int i = 0; i < int(2 * n); ++i) {
    cout << a[i] << ' ';
  }
  cout << '\n';
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}