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 MOD = 1e9 + 7;
const long long INF = LLONG_MAX;
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n, x, flag, m;
    cin >> n;
    x = 2 * n;
    int a[x];
    for (long long i = 0; i < (long long)x; i++) a[i] = i + 1;
    vector<int> s;
    int b[n];
    for (long long i = 0; i < (long long)n; i++) {
      cin >> b[i];
      a[b[i] - 1] = 0;
    }
    for (long long i = 0; i < (long long)n; i++) {
      flag = 0;
      s.push_back(b[i]);
      for (long long j = b[i]; j < (long long)x; j++) {
        if (a[j] > b[i]) {
          flag = 1;
          m = a[j];
          a[j] = 0;
          break;
        }
      }
      if (flag == 1)
        s.push_back(m);
      else
        break;
    }
    if (flag == 0)
      cout << "-1\n";
    else {
      for (long long i = 0; i < (long long)s.size(); i++) cout << s[i] << " ";
      cout << "\n";
    }
  }
  return 0;
}