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;
const long long int mod = 1e9 + 7;
const long long int INF = 1e18;
void solve() {
  long long int n;
  cin >> n;
  long long int a[n], b[2 * n];
  for (long long int i = 0; i < n; i++) cin >> a[i];
  set<long long int> s;
  for (long long int i = 1; i < 2 * n + 1; i++) s.insert(i);
  long long int ind = 0;
  for (long long int i = 0; i < n; i++) {
    s.erase(s.find(a[i]));
  }
  for (long long int i = 0; i < 2 * n; i++) {
    b[i] = a[ind];
    auto it = s.lower_bound(a[ind]);
    if (it == s.end()) {
      cout << -1 << "\n";
      return;
    }
    b[i + 1] = *it;
    s.erase(it);
    i++;
    ind++;
  }
  for (long long int i = 0; i < 2 * n; i++) cout << b[i] << " ";
  cout << "\n";
  return;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long int t = 1;
  cin >> t;
  for (long long int i = 1; i < t + 1; i++) {
    solve();
  }
  return 0;
}