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() {
  long long t;
  cin >> t;
  while (t--) {
    long long n, i, p = 0, q, h, r = 0, j;
    cin >> n;
    long long a[n], b[2 * n], c[2 * n];
    memset(c, 0, sizeof(c));
    for (i = 0; i < n; i++) {
      cin >> a[i];
    }
    for (i = 0; i < n; i++) {
      c[a[i] - 1] = 1;
    }
    for (i = 0; i < n; i++) {
      h = 1;
      b[2 * i] = a[i];
      for (j = a[i]; j < 2 * n; j++) {
        if (c[j] != 1) {
          break;
        }
      }
      b[2 * i + 1] = j + 1;
      if (j < 2 * n) c[j] = 1;
    }
    for (i = 0; i < 2 * n; i++) {
      if (b[i] <= 2 * n) c[b[i] - 1] = 1;
    }
    for (i = 0; i < 2 * n; i++) {
      if (c[i] == 0) {
        r = 1;
        break;
      }
    }
    if (r == 1) {
      cout << "-1" << endl;
    } else {
      for (i = 0; i < 2 * n; i++) {
        cout << b[i] << " ";
      }
      cout << endl;
    }
  }
}