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;
int main() {
  int test;
  cin >> test;
  while (test--) {
    int n;
    cin >> n;
    int a[n + 5];
    int v[2 * n + 10], u[2 * n + 10];
    for (int i = 1; i <= 2 * n; i++) {
      v[i] = i;
      u[i] = i;
    }
    int j = 1;
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
      v[a[i]] = 0;
      u[a[i]] = 0;
    }
    int s = 0;
    for (int i = 1; i <= n; i++) {
      for (int j = a[i]; j <= 2 * n; j++) {
        if (v[j] > a[i]) {
          s++;
          v[j] = 0;
          break;
        }
      }
    }
    if (s != n)
      cout << "-1";
    else {
      for (int i = 1; i <= n; i++) {
        for (int j = a[i]; j <= 2 * n; j++) {
          if (u[j] > a[i]) {
            cout << a[i] << " " << u[j] << " ";
            u[j] = 0;
            break;
          }
        }
      }
    }
    cout << endl;
  }
  return 0;
}