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 
#include <bits/stdc++.h>
using namespace std;
const long int full = 205;
int n;
int b[full], a[full];
bool f[full];
int find(int s) {
  int x = 0;
  for (int i = s + 1; i <= 2 * n; i++)
    if (f[i] == false) {
      f[i] = true;
      x = i;
      break;
    }
  return x;
}
void process() {
  for (int i = 1; i <= 2 * n; i++) f[i] = false;
  for (int i = 1; i <= n; i++) {
    a[2 * i - 1] = b[i];
    f[b[i]] = true;
  }
  int x;
  for (int i = 1; i <= n; i++) {
    x = find(b[i]);
    if (x == 0) {
      cout << "-1" << endl;
      return;
    }
    a[i * 2] = x;
  }
  for (int i = 1; i <= 2 * n; i++) cout << a[i] << " ";
  cout << endl;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  long int t;
  cin >> t;
  while (t--) {
    cin >> n;
    for (int i = 1; i <= n; i++) cin >> b[i];
    process();
  }
  return 0;
}
