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;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<int> b(n + 5);
    vector<int> a(2 * n + 10);
    vector<pair<int, int>> p;
    map<int, int> mp;
    for (int i = 1; i <= n; i++) {
      cin >> b[i];
      mp[b[i]]++;
      a[2 * i - 1] = b[i];
      p.push_back({b[i], 2 * i});
    }
    bool ok = true;
    for (int i = 0; i < p.size(); i++) {
      int st = p[i].first + 1;
      while (st <= 2 * n && mp[st] > 0) {
        st++;
      }
      if (st > 2 * n) {
        ok = false;
        break;
      }
      a[p[i].second] = st;
      mp[st]++;
    }
    if (!ok) {
      cout << -1 << "\n";
      continue;
    }
    for (int i = 1; i <= n; i++) {
      if (a[2 * i - 1] > a[2 * i]) swap(a[2 * i - 1], a[2 * i]);
    }
    map<int, int> kek;
    for (int i = 1; i <= 2 * n; i++) {
      kek[a[i]]++;
    }
    for (int i = 1; i <= 2 * n; i++) {
      if (kek[i] != 1) {
        ok = false;
        break;
      }
    }
    if (!ok) {
      cout << -1 << "\n";
      continue;
    }
    for (int i = 1; i <= 2 * n; i++) {
      cout << a[i] << " ";
    }
    cout << "\n";
  }
  return 0;
}