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 long long int mod = 1e9 + 7;
const double pie = 3.14159265358979323846;
const char nl = '\n';
const char spc = ' ';
template <typename Type>
istream &operator>>(istream &in, vector<Type> &v) {
  long long int n = v.size();
  for (int i = 0; i < n; i++) in >> v[i];
  return in;
}
template <typename Type>
ostream &operator<<(ostream &out, vector<Type> &v) {
  for (auto value : v) out << value << ' ';
  out << nl;
  return out;
}
bool isPrime(long long int n) {
  if (n < 4) return true;
  if (n % 2 == 0) return false;
  long long int i = 3;
  while (i * i <= n) {
    if (n % i == 0) return false;
    i += 2;
  }
  return true;
}
void solve() {
  long long int n;
  cin >> n;
  vector<long long int> b(n);
  vector<long long int> cnt(300, 0);
  for (long long int i = 0; i < n; ++i) {
    cin >> b[i];
    cnt[b[i]] = 1;
  }
  vector<long long int> a;
  for (long long int i = 0; i < n; ++i) {
    a.emplace_back(b[i]);
    long long int k = b[i] + 1;
    while (cnt[k]) ++k;
    a.emplace_back(k);
    cnt[b[i]] = cnt[k] = 1;
  }
  for (long long int i = 1; i <= 2 * n; ++i) {
    if (cnt[i] != 1) {
      cout << "-1" << nl;
      return;
    }
  }
  cout << a;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}