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 MAX_N = 105;
long long t, n, a[MAX_N], b;
vector<int> ans;
set<int> second;
int main() {
  cin >> t;
  while (t--) {
    cin >> n;
    second.clear();
    b = 1;
    ans.resize(2 * n + 2);
    fill(ans.begin(), ans.end(), 0);
    for (int i = 1; i <= 2 * n; ++i) {
      second.insert(i);
    }
    for (int i = 1; i <= n; ++i) {
      cin >> a[i];
      ans[2 * i - 1] = a[i];
      second.erase(a[i]);
    }
    for (int i = 1; i < 2 * n; i += 2) {
      auto it = second.lower_bound(ans[i]);
      if (it == second.end()) {
        b = false;
        break;
      }
      ans[i + 1] = *it;
      second.erase(it);
    }
    if (!b) {
      cout << -1 << endl;
      continue;
    }
    for (int i = 1; i < ans.size() - 1; ++i) {
      cout << ans[i] << " ";
    }
    cout << endl;
  }
  return 0;
}