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 = 998244353;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    int arr[n];
    unordered_set<int> us;
    for (int i = 0; i < n; i++) {
      cin >> arr[i];
      us.insert(arr[i]);
    }
    bool flag = false;
    multiset<int> m;
    for (int i = 1; i <= 2 * n; i++) {
      if (us.find(i) == us.end()) {
        m.insert(i);
      }
    }
    int ans[2 * n + 1];
    for (int i = 1; i <= n; i++) {
      int temp = arr[i - 1];
      if (m.upper_bound(temp) == m.end()) {
        flag = true;
        break;
      } else {
        auto y = m.upper_bound(temp);
        ans[2 * i - 1] = temp;
        ans[2 * i] = (*y);
        m.erase(*y);
      }
    }
    if (flag == true) {
      cout << -1 << "\n";
    } else {
      for (int i = 1; i <= 2 * n; i++) {
        cout << ans[i] << " ";
      }
      cout << "\n";
    }
  }
  return 0;
}