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;
long long int n, m;
long long int arr[3000000];
long long int brr[3000000];
long long int an[3000000];
array<long long int, 2> dp[3000000];
string st;
vector<long long int> vec;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int TESTS, q, a, s, b, r, k, c, p, w, d, x, i, j, y, z, xs, ys, t;
  TESTS = 1;
  cin >> TESTS;
  while (TESTS--) {
    cin >> n;
    for (long long int i = 1; i <= 2 * n; i++) brr[i] = 0;
    for (long long int i = 0; i <= n - 1; i++) {
      cin >> arr[i];
      brr[arr[i]] = 1;
    }
    bool pos = true;
    for (long long int i = 0; i <= n - 1; i++) {
      an[2 * i] = arr[i];
      bool cur = false;
      for (long long int j = arr[i]; j <= 2 * n; j++)
        if (brr[j] == 0) {
          cur = true;
          an[2 * i + 1] = j;
          brr[j] = 1;
          break;
        }
      if (!cur) {
        pos = false;
        break;
      }
    }
    if (!pos) {
      cout << -1 << "\n";
      continue;
    }
    for (long long int i = 0; i <= 2 * n - 1; i++) cout << an[i] << " ";
    cout << "\n";
  }
  return 0;
}