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>
template <typename T>
std::istream& operator>>(std::istream& input, std::vector<T>& v) {
  for (T& a : v) input >> a;
  return input;
}
void no_answer() { std::cout << -1 << '\n'; }
void answer(const std::vector<unsigned>& v) {
  const char* separator = "";
  for (const unsigned x : v) {
    std::cout << separator << x;
    separator = " ";
  }
  std::cout << '\n';
}
void solve(const std::vector<unsigned>& b) {
  const size_t n = b.size();
  std::set<unsigned> s;
  for (unsigned x = 1; x <= 2 * n; ++x) s.insert(x);
  for (const unsigned x : b) s.erase(x);
  std::vector<unsigned> a(2 * n);
  for (size_t i = 0; i < n; ++i) {
    const auto it = s.upper_bound(b[i]);
    if (it == s.cend()) return no_answer();
    a[2 * i + 0] = b[i];
    a[2 * i + 1] = *it;
    s.erase(it);
  }
  answer(a);
}
void test_case() {
  size_t n;
  std::cin >> n;
  std::vector<unsigned> b(n);
  std::cin >> b;
  solve(b);
}
int main() {
  size_t t;
  std::cin >> t;
  while (t-- > 0) test_case();
  return 0;
}