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 faster() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  if (0) {
    freopen(
        "input"
        "sum.in",
        "r", stdin);
    freopen(
        "input"
        "sum.out",
        "w", stdout);
  }
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
long long fact(int n) {
  long long f = 1;
  if ((n == 0) || (n == 1))
    f = 1;
  else
    for (int i = 1; i <= n; i++) f *= i;
  return f;
}
int fib(int n) {
  int a = 0, b = 1;
  for (int i = 0; i < n; ++i) {
    b = b + a;
    a = b - a;
  }
  return b;
}
int n;
void solve() {
  cin >> n;
  bool b = true;
  std::vector<int> v(n);
  for (int i = 0; i < n; ++i) cin >> v[i];
  std::vector<int> v1;
  for (int i = 1; i <= 2 * n; ++i) {
    if (find(v.begin(), v.end(), i) == v.end()) v1.push_back(i);
  }
  sort(v1.begin(), v1.end());
  std::vector<int> v2(2 * n);
  for (int i = 0; i < n; ++i) {
    v2[2 * i] = v[i];
    auto kos = upper_bound(v1.begin(), v1.end(), v[i]);
    if (kos == v1.end()) {
      b = false;
      break;
    } else {
      v2[2 * i + 1] = *kos;
      v1.erase(kos);
    }
  }
  if (!b)
    cout << -1;
  else {
    for (int i = 0; i < 2 * n; ++i) cout << v2[i] << ' ';
  }
}
int main() {
  faster();
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
    cout << "\n";
  }
  return 0;
}