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>
long long MOD = 1000000007;
using namespace std;
bool sortbysec(const pair<long long, long long> &a,
               const pair<long long, long long> &b) {
  return (a.second < b.second);
}
long long POW(long long a, long long b) {
  if (b == 0) return 1;
  if (b == 1) return a % MOD;
  long long temp = POW(a, b / 2);
  if (b % 2 == 0)
    return (temp * temp) % MOD;
  else
    return (((temp * temp) % MOD) * a) % MOD;
}
vector<char> capl = {'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I',
                     'J', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R',
                     'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z'};
vector<char> sml = {'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i',
                    'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r',
                    's', 't', 'u', 'v', 'w', 'x', 'y', 'z'};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    int arr[n];
    map<int, int> m;
    for (int i = 1; i < 201; i++) {
      m[i]++;
    }
    for (int i = 0; i < n; i++) {
      cin >> arr[i];
      m[arr[i]]--;
    }
    int b[2 * n];
    for (int i = 0; i < n; i++) {
      b[2 * i] = arr[i];
      for (auto x : m) {
        if (x.first > arr[i] && x.second != 0) {
          b[2 * i + 1] = x.first;
          m[x.first] = 0;
          break;
        }
      }
    }
    int c[2 * n], f = 0;
    copy(b, b + 2 * n, c);
    sort(c, c + 2 * n);
    for (int i = 0; i < 2 * n; i++) {
      if (c[i] != i + 1) {
        f = 1;
        break;
      }
    }
    if (f == 1) {
      cout << "-1\n";
    } else {
      for (int i = 0; i < 2 * n; i++) {
        cout << b[i] << " ";
      }
      cout << "\n";
    }
  }
}