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() {
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    int a[n];
    vector<bool> flag(2 * n + 1, false);
    for (int i = 0; i < n; i++) {
      cin >> a[i];
      flag[a[i]] = true;
    }
    vector<int> st;
    for (int i = 1; i <= 2 * n; i++) {
      if (!flag[i]) {
        st.push_back(i);
      }
    }
    if (flag[2 * n] || !flag[1]) {
      cout << -1 << endl;
    } else {
      vector<int> b;
      int d = 0;
      for (int i = 0; i < n; i++) {
        b.push_back(a[i]);
        vector<int>::iterator itr;
        for (itr = st.begin(); itr != st.end(); itr++) {
          if (*itr > a[i]) {
            break;
          }
        }
        if (itr == st.end()) {
          cout << -1 << endl;
          d++;
          break;
        }
        b.push_back(*itr);
        st.erase(itr);
      }
      if (d) {
        continue;
      }
      for (int i = 0; i < b.size(); i++) {
        cout << b[i] << " ";
      }
      cout << endl;
    }
  }
  return 0;
}