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>
#pragma GCC optimize("O2")
using namespace std;
signed main() {
  long long nn;
  cin >> nn;
  while (nn--) {
    long long n;
    cin >> n;
    vector<long long> vec(n + 1);
    vector<long long> res(2 * n + 1);
    vector<long long> vissited(2 * n + 1, 0);
    for (long long i = 1; i <= n; ++i) {
      cin >> vec[i];
      res[2 * i - 1] = vec[i];
      vissited[vec[i]] = 1;
    }
    bool ctr = 0;
    for (long long j = 1; j <= n; ++j) {
      long long cur = vec[j] + 1;
      while (cur <= 2 * n and vissited[cur] == 1) cur++;
      if (cur == 2 * n + 1) {
        cout << -1 << "\n";
        ctr = 1;
        break;
      }
      vissited[cur] = 1;
      res[2 * j] = cur;
    }
    if (ctr) continue;
    for (long long k = 1; k <= 2 * n; ++k) {
      cout << res[k] << " ";
    }
    cout << "\n";
  }
  return 0;
}