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 N = 410;
int b[N];
int a[N];
int t, n;
bool used[N];
int find(int x) {
  for (int i = x + 1; i <= 2 * n; i++) {
    if (!used[i]) {
      return i;
    }
  }
  return -1;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> t;
  while (t--) {
    memset(used, 0, sizeof(used));
    cin >> n;
    for (int i = 1; i <= n; i++) {
      cin >> b[i];
      used[b[i]] = 1;
    }
    bool noans = 0;
    for (int i = 1; i <= n; i++) {
      int f = find(b[i]);
      if (f == -1) {
        noans = 1;
        cout << -1 << endl;
        break;
      } else {
        a[2 * i - 1] = b[i];
        a[2 * i] = f;
        used[f] = 1;
      }
    }
    if (!noans) {
      for (int i = 1; i <= 2 * n; i++) {
        cout << a[i] << " ";
      }
      cout << endl;
    }
  }
  return 0;
}