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;
long long t, n, x[1005000];
void solve() {
  cin >> n;
  set<long long> s;
  vector<pair<long long, long long> > b;
  vector<pair<long long, pair<long long, long long> > > v;
  for (int i = 1; i <= 2 * n; i++) s.insert(i);
  for (int i = 1; i <= n; i++)
    cin >> x[i], s.erase(x[i]), b.push_back({x[i], i});
  vector<long long> a((s).begin(), (s).end());
  sort((b).begin(), (b).end());
  for (int i = 0; i < n; i++) {
    if (a[i] <= b[i].first) {
      cout << -1 << endl;
      return;
    }
  }
  for (int i = 0; i < n; i++) swap(b[i].first, b[i].second);
  sort((b).begin(), (b).end());
  for (int i = 0; i < n; i++) {
    cout << b[i].second << ' ' << *(s.upper_bound(b[i].second)) << ' ';
    s.erase(s.upper_bound(b[i].second));
  }
  cout << endl;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> t;
  while (t--) solve();
  return 0;
}