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 long long M = 1e8 + 7, maxn = 4e5 + 5;
long long int ans, n, m, x, y, q, k, a, b, sum;
int32_t main() {
  cin >> q;
  while (q--) {
    cin >> n;
    long long int a[n];
    set<long long int> s;
    k = 0;
    for (long long int i = 1; i <= 2 * n; i++) s.insert(i);
    s.insert(M);
    for (long long int i = 0; i < n; i++) {
      cin >> a[i];
      if (!s.count(a[i])) {
        k = 1;
      } else
        s.erase(a[i]);
    }
    long long int b[2 * n];
    long long int j = 0;
    for (long long int i = 1; i < 2 * n; i += 2) {
      x = *s.lower_bound(a[j]);
      if (x == M) {
        k = 1;
        break;
      }
      b[i] = x;
      s.erase(x);
      j++;
    }
    j = 0;
    if (k) {
      cout << -1 << endl;
    } else {
      for (long long int i = 1; i < 2 * n; i += 2) {
        cout << a[j] << " " << b[i] << " ";
        j++;
      }
      cout << endl;
    }
  }
  return 0;
}