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 t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    int a[n];
    int maxi = 2 * n;
    unordered_map<int, int> map;
    for (long long i = 0; i < n; i++) {
      cin >> a[i];
      map[a[i]]++;
    }
    if (map.count(1) <= 0) {
      cout << -1 << endl;
      continue;
    }
    if (map.count(maxi) > 0) {
      cout << -1 << endl;
      continue;
    }
    vector<int> ans;
    for (int i = 0; i < n; i++) {
      int greater = -1;
      for (int j = a[i] + 1; j <= 2 * n; j++) {
        if (map.count(j) <= 0) {
          map[j]++;
          greater = j;
          break;
        }
      }
      if (greater == -1) {
        cout << -1;
        break;
      } else {
        ans.push_back(a[i]);
        ;
        ans.push_back(greater);
        ;
      }
    }
    if (ans.size() == 2 * n) {
      for (long long i = 0; i < ans.size(); i++) {
        cout << ans[i] << " ";
      }
    }
    cout << endl;
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve();
}