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;
void solve() {
  long long n;
  cin >> n;
  long long a[n + 10];
  map<long long, long long> mp;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    mp[a[i]] = 1;
  }
  vector<long long> v;
  for (long long i = 0; i < n; i++) {
    v.push_back(a[i]);
    for (long long j = a[i] + 1; j <= 2 * n; j++) {
      if (mp[j] == 0) {
        mp[j] = 1;
        v.push_back(j);
        break;
      }
    }
  }
  if (v.size() == 2 * n) {
    for (auto x : v) {
      cout << x << " ";
    }
    cout << "\n";
    return;
  }
  cout << "-1"
       << "\n";
}
int main() {
  long long t;
  cin >> t;
  while (t--) solve();
}