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 main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    set<int> s;
    for (int i = 1; i <= 2 * n; i++) s.insert(i);
    vector<int> b(n + 1);
    vector<int> a(2 * n + 1, 0);
    for (int i = 1; i <= n; i++) cin >> b[i];
    int flag = 0;
    for (int i = 1; i <= n; i++) {
      if (s.find(b[i]) != s.end()) {
        s.erase(b[i]);
        a[2 * i - 1] = b[i];
      } else {
        flag = 1;
        break;
      }
    }
    if (flag == 0) {
      set<int>::iterator it;
      for (it = s.begin(); it != s.end() && flag == 0; it++) {
        int flag1 = 0;
        for (int i = 2; i <= 2 * n; i = i + 2) {
          if (a[i] == 0) {
            if (a[i - 1] < *it) {
              a[i] = *it, flag1 = 1;
              break;
            }
          }
        }
        if (flag1 == 0) flag = 1;
      }
    }
    if (flag == 1)
      cout << -1;
    else {
      for (int i = 1; i <= 2 * n; i++) cout << a[i] << " ";
    }
    cout << endl;
  }
  return 0;
}