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;
bool present[1000] = {0};
int main(int argc, char **argv) {
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    int arr[n];
    for (int i = 0; i < n; i++) cin >> arr[i];
    for (int i = 1; i <= 2 * n; i++) present[i] = false;
    for (int i = 0; i < n; i++) present[arr[i]] = true;
    vector<int> choose;
    for (int i = 1; i <= 2 * n; i++)
      if (!present[i]) choose.push_back(i);
    vector<int> result;
    bool possible = true;
    for (int i = 0; i < n; i++) {
      auto idx = lower_bound(choose.begin(), choose.end(), arr[i]);
      if (idx == choose.end()) {
        possible = false;
      } else {
        result.push_back(*idx);
        choose.erase(idx);
      }
    }
    if (!possible) {
      cout << -1 << endl;
    } else {
      for (int i = 0; i < n; i++) cout << arr[i] << " " << result[i] << " ";
      cout << endl;
    }
  }
  return 0;
}