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 N = 1e6 + 7;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    long long int a[n];
    map<long long int, long long int> m;
    for (int i = 0; i < n; i++) {
      cin >> a[i];
      m[a[i]] = 1;
    }
    vector<long long int> v;
    int yo = 0;
    for (int i = 0; i < n; i++) {
      if (a[i] + 1 > 2 * n) {
        yo = 1;
        break;
      }
      if (!m[a[i] + 1]) {
        v.push_back(a[i]);
        v.push_back(a[i] + 1);
        m[a[i] + 1] = 1;
      } else {
        long long int z = -1;
        for (int j = a[i] + 1; j <= 2 * n; j++) {
          if (!m[j]) {
            z = j;
            break;
          }
        }
        if (z != -1) {
          v.push_back(a[i]);
          v.push_back(z);
          m[z] = 1;
        } else
          yo = 1;
      }
      if (yo == 1) break;
    }
    if (yo == 1)
      cout << "-1" << endl;
    else {
      for (int i = 0; i < v.size(); i++) cout << v[i] << " ";
      cout << endl;
    }
  }
  return 0;
}