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;
map<int, int> m1;
int a[101];
int b[101];
int main() {
  int t, n, i, k, key;
  scanf("%d", &t);
  while (t--) {
    m1.clear();
    scanf("%d", &n);
    for (i = 1; i <= n; i++) {
      scanf("%d", &b[i]);
      m1[b[i]] = 1;
    }
    key = 1;
    for (i = 1; i <= n; i++) {
      k = b[i];
      while (m1.count(k)) {
        k++;
      }
      if (k <= 2 * n) {
        a[i] = k;
        m1[a[i]] = 1;
      } else {
        key = 0;
        break;
      }
    }
    if (key) {
      for (i = 1; i <= n; i++) {
        if (i == n) {
          printf("%d %d\n", b[i], a[i]);
        } else {
          printf("%d %d ", b[i], a[i]);
        }
      }
    } else {
      printf("-1\n");
    }
  }
  return 0;
}