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 int N = 3e5 + 500;
const long long mod = 1e9 + 7;
const long long cmod = 998244353;
const long long inf = 1LL << 61;
const int M = 1e6 + 500;
const long long int ths = 1LL << 40;
const int NN = 5e3 + 6;
void solve() {
  long long int n;
  cin >> n;
  long long int b[n + 1], a[2 * n + 1];
  set<long long int> s;
  for (int i = 1; i <= 2 * n; i++) s.insert(i);
  for (int i = 1; i <= n; i++) {
    cin >> b[i];
    s.erase(b[i]);
  }
  vector<long long int> v;
  for (auto it : s) v.push_back(it);
  long long int idx = 1;
  bool ok = 1;
  for (int i = 1; i <= n; i++) {
    a[idx] = b[i];
    idx++;
    long long int x = *s.upper_bound(b[i]);
    if (s.count(x) == 0) {
      ok = 0;
      break;
    }
    s.erase(x);
    a[idx] = x;
    idx++;
  }
  if (!ok) {
    cout << -1;
  } else {
    for (int i = 1; i <= 2 * n; i++) cout << a[i] << ' ';
  }
  cout << '\n';
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}