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;
bool sortbysec(const pair<int, int> &a, const pair<int, int> &b) {
  return (a.second < b.second);
}
void solve() {
  int n;
  cin >> n;
  vector<pair<int, int>> m;
  map<int, int> mp;
  for (int j = 0; j < n; j++) {
    int x;
    cin >> x;
    m.push_back({x, j * 2});
    mp.insert({x, j * 2});
  }
  set<int> v;
  for (int i = 1; i < 2 * n + 1; i++) {
    if (mp.find(i) == mp.end()) {
      v.insert(i);
    }
  }
  for (int i = 0; i < n; i++) {
    auto it = v.upper_bound(m[i].first);
    m.push_back({*it, m[i].second + 1});
    v.erase(*it);
  }
  sort(m.begin(), m.end(), sortbysec);
  for (int i = 0; i < n; i++) {
    if (m[2 * i + 1].first <= m[2 * i].first) {
      cout << "-1\n";
      return;
    }
  }
  for (auto u : m) {
    cout << u.first << " ";
  }
  cout << "\n";
}
int main() {
  int t;
  cin >> t;
  for (int i = 0; i < t; i++) {
    solve();
  }
}