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;
bool compare(const pair<string, string>& i, const pair<string, string>& j) {
  if (i.first != j.first)
    return i.first < j.first;
  else
    return i.second < j.second;
}
vector<long long> factorize(long long n) {
  vector<long long> v;
  for (long long i = 2; i * i <= n; i++) {
    while (n % i == 0) {
      v.push_back(i);
      n /= i;
    }
  }
  if (n != 1) v.push_back(n);
  return v;
}
bool isPallindrome(string s) {
  for (long long i = 0; i <= s.length() / 2; i++) {
    if (s[i] != s[s.length() - i - 1]) return false;
  }
  return true;
}
long long GCD(long long a, long long b) {
  if (b == 0) return a;
  return GCD(b, a % b);
}
vector<bool> sieve(long long n) {
  vector<bool> PRIME(n + 1, true);
  PRIME[0] = false;
  PRIME[1] = false;
  for (long long p = 2; p * p <= n; p++) {
    if (PRIME[p] == true) {
      for (long long i = p * 2; i <= n; i += p) {
        PRIME[i] = false;
      }
    }
  }
  return PRIME;
}
bool isPrime(long long n) {
  if (n == 1) return false;
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) return false;
  }
  return true;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  long long t;
  cin >> t;
  while (t--) {
    long long n, a;
    cin >> n;
    vector<long long> vec;
    vector<bool> v((2 * n) + 1, false);
    v[0] = true;
    for (long long i = 0; i < n; i++) {
      cin >> a;
      v[a] = true;
      vec.push_back(a);
    }
    for (long long i = 0; i < n; i++) {
      long long p = vec[i] + 1;
      while (true) {
        if (p > 2 * n) {
          cout << "-1"
               << "\n";
          goto h;
        }
        if (v[p] == false) {
          vec.push_back(p);
          v[p] = true;
          break;
        }
        p++;
      }
    }
    if (count((v).begin(), (v).end(), false) != 0)
      cout << "-1"
           << "\n";
    else {
      for (long long i = 0; i < n; i++)
        cout << vec[i] << " " << vec[i + n] << " ";
      cout << "\n";
    }
  h:;
  }
}