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;
set<int> prime;
void SieveOfEratosthenes(int n) {
  bool prme[n + 1];
  memset(prme, true, sizeof(prme));
  for (int p = 2; p * p <= n; p++) {
    if (prme[p] == true) {
      for (int i = p * p; i <= n; i += p) prme[i] = false;
    }
  }
  for (int p = 2; p <= n; p++)
    if (prme[p]) prime.insert(p);
}
void solve() {
  int n;
  cin >> n;
  vector<int> in(n), v;
  set<int> s, temp;
  for (int i = 0; i < n; i++) {
    cin >> in[i];
    s.insert(in[i]);
  }
  for (auto i : in) {
    v.push_back(i);
    int x = i + 1;
    while (s.count(x) || temp.count(x)) {
      x++;
    }
    v.push_back(x);
    temp.insert(x);
  }
  auto it = temp.end();
  it--;
  if ((*it) != (2 * n)) {
    cout << -1;
  } else {
    for (auto i : v) cout << i << ' ';
  }
  cout << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) solve();
}