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 INF = 0x3f3f3f3f;
const double PI = acos(-1.0);
const int T = 102;
int a[T], b[T];
int main() {
  ios_base::sync_with_stdio(false);
  int tc;
  cin >> tc;
  while (tc--) {
    int n;
    cin >> n;
    vector<pair<int, int> > need;
    set<int> disp;
    for (int i = 1; i <= 2 * n; i++) disp.insert(i);
    for (int i = 1; i <= n; i++) {
      cin >> b[i];
      disp.erase(b[i]);
      a[2 * i - 1] = b[i];
      need.emplace_back(2 * i, b[i]);
    }
    sort(need.begin(), need.end(), greater<pair<int, int> >());
    bool f = 0;
    while (need.size()) {
      pair<int, int> top = need.back();
      need.pop_back();
      int at = -1;
      for (auto x : disp) {
        if (x > top.second) {
          at = x;
          break;
        }
      }
      if (at == -1) {
        f = 1;
        break;
      } else {
        disp.erase(at);
        a[top.first] = at;
      }
    }
    if (f)
      cout << -1 << '\n';
    else {
      for (int i = 1; i <= 2 * n; i++) cout << a[i] << " ";
      cout << '\n';
    }
  }
  return 0;
}