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;
int main() {
  long long int tc;
  cin >> tc;
  while (tc--) {
    long long int n;
    cin >> n;
    long long int arr[2 * n + 1];
    long long int e = 0;
    vector<long long int> v;
    long long int ee = 0;
    for (long long int i = 1; i < n + 1; i++) {
      cin >> arr[2 * i - 1];
      if (arr[2 * i - 1] > 2 * n) e = 1;
      v.push_back(arr[2 * i - 1]);
    }
    if (e == 1) {
      cout << -1 << endl;
      continue;
    }
    sort(v.begin(), v.end());
    vector<long long int> v1;
    for (long long int i = 1; i < 2 * n + 1; i++) {
      long long int b = lower_bound(v.begin(), v.end(), i) - v.begin();
      if (v[b] == i)
        continue;
      else
        v1.push_back(i);
    }
    sort(v1.begin(), v1.end());
    for (long long int i = 1; i < n + 1; i++) {
      long long int bb =
          lower_bound(v1.begin(), v1.end(), arr[2 * i - 1]) - v1.begin();
      if (bb >= v1.size()) {
        ee = 1;
        break;
      } else {
        arr[2 * i] = v1[bb];
        v1.erase(bb + v1.begin());
      }
    }
    if (ee == 1)
      cout << -1 << endl;
    else {
      for (long long int i = 1; i < 2 * n + 1; i++) cout << arr[i] << " ";
      cout << endl;
    }
  }
  return 0;
}