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;
int arr[105];
unordered_set<int> used;
bool fl = 1;
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    for (int i = 0; i < n; ++i) {
      cin >> arr[i];
      used.insert(arr[i]);
    }
    vector<int> v;
    for (int i = 0; i < n; ++i) {
      v.push_back(arr[i]);
      int j = arr[i] + 1;
      while (used.find(j) != used.end()) ++j;
      if (j > 2 * n)
        break;
      else {
        v.push_back(j);
      }
      used.insert(j);
    }
    if (v.size() == 2 * n)
      for (int i = 0; i < v.size(); ++i) cout << v[i] << ' ';
    else
      cout << -1;
    cout << '\n';
    v.clear();
    used.clear();
  }
}