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;
void solve(int n, int b[]) {
  set<int> s;
  int a[2 * n];
  for (int i = 0; i < (int)2 * n; i++) {
    s.insert(i + 1);
  }
  for (int i = 0; i < (int)n; i++) {
    s.erase(b[i]);
  }
  if (*min_element(s.begin(), s.end()) == 1 ||
      *max_element(b, b + n) == 2 * n) {
    cout << "-1\n";
    return;
  } else {
    for (int i = 0; i < (int)2 * n; i++) {
      if (i % 2 == 0)
        a[i] = b[i / 2];
      else {
        if ((s.upper_bound(b[i / 2])) != s.end()) {
          a[i] = *s.upper_bound(b[i / 2]);
          s.erase(*s.upper_bound(b[i / 2]));
        } else {
          cout << "-1\n";
          return;
        }
      }
    }
  }
  for (auto x : a) cout << x << " ";
  cout << "\n";
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int tc;
  cin >> tc;
  while (tc--) {
    int n;
    cin >> n;
    int b[n];
    for (int i = 0; i < (int)n; i++) {
      cin >> b[i];
    }
    solve(n, b);
  }
  return 0;
}