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;
long long t, n, a[1005], dd[2005], kt, b[2005], m;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> t;
  while (t--) {
    memset(dd, 0, sizeof(dd));
    memset(b, 0, sizeof(b));
    m = 0;
    cin >> n;
    for (long long i = 1; i <= n; i++) {
      cin >> a[i];
      dd[a[i]] = 1;
    }
    for (long long i = 1; i <= n; i++) {
      kt = 0;
      for (long long j = a[i] + 1; j <= 2 * n; j++)
        if (dd[j] == 0) {
          kt = j;
          break;
        }
      if (kt == 0) {
        m = -1;
        break;
      } else {
        m++;
        b[m] = a[i];
        m++;
        b[m] = kt;
        dd[kt] = 1;
      }
    }
    if (m == -1)
      cout << m << '\n';
    else {
      for (long long i = 1; i <= m; i++) cout << b[i] << " ";
      cout << '\n';
    }
  }
  return 0;
}