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 int N = 1e5;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int tc, ca = 0;
  cin >> tc;
  while (tc--) {
    int n, el;
    vector<int> b;
    vector<int> baki;
    map<int, int> m1;
    map<int, int> m2;
    cin >> n;
    for (int i = 0; i < n; i++) {
      cin >> el;
      m1[el] = 1;
      b.push_back(el);
    }
    for (int i = 1; i <= n + n; i++) {
      if (!m1[i]) baki.push_back(i);
    }
    m1.clear();
    bool ok = false;
    vector<int> a;
    for (int i = 0; i < n; i++) {
      int p1 = b[i];
      for (int j = 0; j < n; j++) {
        if (baki[j] && p1 <= baki[j]) {
          a.push_back(p1);
          a.push_back(baki[j]);
          baki[j] = 0;
          break;
        }
      }
    }
    if (a.size() != n + n)
      cout << "-1" << '\n';
    else {
      for (int i = 0; i < n + n; i++) {
        cout << a[i];
        if (i != (n + n) - 1) cout << ' ';
      }
      cout << '\n';
    }
  }
  return 0;
}