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 int mod = 998244353.;
const int N = 1e6 + 5;
bool sortbysec(const pair<pair<long long, long long>, long long> &a,
               const pair<pair<long long, long long>, long long> &b) {
  return (a.second > b.second);
}
long long powermod(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
bool isprime(int n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (int i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long t;
  cin >> t;
loop:
  while (t--) {
    long long n;
    cin >> n;
    vector<long long> a(n);
    set<long long> s;
    for (long long i = 0; i < n; i++) {
      cin >> a[i];
      s.insert(a[i]);
    }
    vector<long long> b;
    for (long long i = 0; i < n; i++) {
      b.push_back(a[i]);
      s.insert(a[i]);
      long long f = 0;
      for (long long j = a[i] + 1; j <= 2 * n; j++) {
        if (s.find(j) == s.end()) {
          f = 1;
          s.insert(j);
          b.push_back(j);
          break;
        }
      }
      if (f == 0) {
        cout << -1 << "\n";
        goto loop;
      }
    }
    for (long long i = 0; i < b.size(); i++) cout << b[i] << " ";
    ;
    cout << "\n";
  }
  return 0;
}