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>
int main() {
  int i, j, k, t, n, l, largest_element;
  scanf("%d", &t);
  int result[t][200];
  int largest_element_array[t];
  for (i = 0; i < t; i++) {
    int count = 0;
    scanf("%d", &n);
    largest_element = 2 * n;
    largest_element_array[i] = largest_element;
    int array_given[n];
    for (j = 0; j < n; j++) {
      scanf("%d", &array_given[j]);
    }
    int smallest_element = 100000000;
    for (j = 0; j < n; j++) {
      if (array_given[j] < smallest_element) {
        smallest_element = array_given[j];
      }
    }
    for (j = 0; j < n; j++) {
      if (array_given[j] >= largest_element) {
        result[i][0] = -1;
        result[i][1] = 0;
        count += 1;
        break;
      }
    }
    int n_r_elements = 0;
    if (!count) {
      int array_remaining[largest_element];
      for (k = smallest_element + 1; k <= largest_element; k++) {
        int exists = 0;
        for (l = 0; l < n; l++) {
          if (k == array_given[l]) {
            exists = 1;
            break;
          }
        }
        if (!exists) {
          array_remaining[n_r_elements] = k;
          n_r_elements += 1;
        }
      }
      if (n_r_elements < n) {
        result[i][0] = -1;
        result[i][1] = 0;
        count += 1;
      } else {
        int array[n];
        for (k = 0; k < n; k++) {
          array[k] = 1;
        }
        for (k = 0; k < n; k++) {
          result[i][2 * k] = array_given[k];
        }
        for (k = 0; k < n; k++) {
          count = 0;
          result[i][2 * k + 1] = 0;
          for (l = 0; l < n; l++) {
            if (array[l] && (array_remaining[l] > array_given[k])) {
              result[i][2 * k + 1] = array_remaining[l];
              count += 1;
            }
            if (count) {
              array[l] = 0;
              break;
            }
          }
        }
        for (k = 0; k < n; k++) {
          if (result[i][2 * k + 1] == 0) {
            result[i][0] = -1;
            result[i][1] = 0;
          }
        }
      }
    }
  }
  for (i = 0; i < t; i++) {
    for (j = 0; (result[i][j] && (j < largest_element_array[i])); j++) {
      printf("%d ", result[i][j]);
    }
    printf("\n");
  }
  return 0;
}