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;
typedef long long ll;
typedef unsigned long long ul;
int dx[] = {+1, -1, 0, 0};
int dy[] = {0, 0, +1, -1};
vector<long long> vv;
const long long range = 1e7 + 123;
bitset<range> prime;
void sieve() {
  prime[0] = 1;
  prime[1] = 1;
  for (long long i = 2; i <= range; i++) {
    if (prime[i] == 0) {
      for (long long j = i * i; j <= range; j += i) prime[j] = 1;
    }
  }
  vv.push_back(2);
  for (long long i = 3; i <= range; i += 2)
    if (prime[i] == 0) vv.push_back(i);
}
void Main() {
  int n;
  cin >> n;
  map<int, int> mp;
  long long ar[n];
  for (int i = 0; i < n; i++) {
    cin >> ar[i];
    mp[ar[i]]++;
  }
  bool bol = true;
  vector<long long> v;
  for (int i = 0; i < n; i++) {
    v.push_back(ar[i]);
    bool hi = false;
    for (int j = ar[i]; j <= 2 * n; j++) {
      if (mp[j] == 0) {
        v.push_back(j);
        hi = true;
        mp[j]++;
        break;
      }
    }
    if (hi == false) {
      bol = false;
      break;
    }
  }
  if (bol) {
    for (auto u : v) cout << u << " ";
    cout << endl;
  } else
    cout << -1 << endl;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  long long ttt;
  cin >> ttt;
  while (ttt--) {
    Main();
  }
}