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 MAXN = 200009;
const long long MOD = 119 << 23 | 1;
class {
 public:
  int a[222], b[222];
  void solve() {
    int t;
    cin >> t;
    while (t--) {
      int n;
      cin >> n;
      set<int> Se;
      for (int i = 1; i <= 2 * n; ++i) {
        Se.insert(i);
      }
      for (int i = 0; i < n; ++i) cin >> a[i], Se.erase(a[i]);
      bool ok = 1;
      for (int i = 0; i < 2 * n; i += 2) {
        auto it = Se.lower_bound(a[i / 2]);
        if (it == Se.end()) {
          ok = 0;
          break;
        }
        b[i] = a[i / 2];
        b[i + 1] = *it;
        Se.erase(it);
      }
      if (ok) {
        for (int i = 0; i < 2 * n; ++i) cout << b[i] << " ";
        cout << '\n';
      } else {
        cout << -1 << '\n';
      }
    }
  }
} NSPACE;
int main() {
  ;
  ios_base::sync_with_stdio(false);
  cout.tie(0);
  cin.tie(0);
  NSPACE.solve();
}