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 long long mod = 1000000007;
const long long inf = 1000000000;
const long long N = 2 * (1e2) + 5;
long long ar[N];
long long ans[N];
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  long long t = 1;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    vector<long long> done(2 * n + 1, 0);
    for (long long i = 1; i <= n; ++i) {
      cin >> ar[i];
      done[ar[i]] = 1;
    }
    long long d = 1;
    for (long long i = 1; i <= n; ++i) {
      long long x = ar[i];
      long long y = 0;
      for (long long val = x + 1; val <= 2 * n; ++val) {
        if (!done[val]) {
          y = val;
          done[val] = 1;
          break;
        }
      }
      if (!y) {
        d = 0;
        break;
      }
      ans[i] = y;
    }
    if (!d)
      cout << -1 << "\n";
    else {
      for (long long i = 1; i <= n; ++i) {
        cout << ar[i] << " " << ans[i] << " ";
      }
      cout << "\n";
    }
  }
}