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 n, a[209], vis[209], b[109];
set<int> s;
int main() {
  cin.tie(0);
  cout.tie(0);
  ios::sync_with_stdio(0);
  ios_base::sync_with_stdio(0);
  int t;
  cin >> t;
  while (t--) {
    cin >> n;
    s.clear();
    for (int i = 0; i < n * 2; i++) s.insert(i + 1);
    for (int i = 0; i < n; i++) {
      cin >> b[i];
      s.erase(b[i]);
    }
    for (int i = 0; i < n * 2; i += 2) a[i] = b[i / 2];
    bool no = 0;
    memset(vis, 0, sizeof(vis));
    for (auto it = s.begin(); it != s.end(); it++) {
      int b = -1, x = *it;
      for (int i = 0; i < n * 2; i += 2) {
        if (a[i] < x && b == -1 && vis[i] == 0) b = i;
      }
      a[b + 1] = x;
      if (b == -1) no = 1;
      if (no) break;
      vis[b] = 1;
    }
    if (no)
      cout << -1;
    else
      for (int i = 0; i < 2 * n; i++) cout << a[i] << " ";
    cout << "\n";
  }
}