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 int siz = 1e5 + 7;
const int siz2 = 2e5 + 7;
int t, n;
int a[222];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> t;
A:
  while (t--) {
    cin >> n;
    map<int, int> mp;
    vector<int> Ans;
    for (int i = 0; i < n; ++i) {
      cin >> a[i];
      mp[a[i]]++;
    }
    if (n > 1) {
      for (int i = 0; i < n; ++i) {
        for (int j = a[i]; j <= 2 * n; ++j) {
          if (!mp[j]) {
            mp[j]++;
            Ans.push_back(a[i]);
            Ans.push_back(j);
            break;
          }
        }
      }
      if (Ans.size() != 2 * n) {
        cout << -1 << endl;
        goto A;
      }
    } else {
      if (a[0] == 1) {
        cout << 1 << " " << 2 << endl;
        goto A;
      } else {
        cout << -1 << endl;
        goto A;
      }
    }
    for (auto it : Ans) cout << it << " ";
    cout << endl;
  }
  return 0;
}