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 M = 220;
int a[M], v[M];
int t, n;
int main() {
  scanf("%d", &t);
  while (t--) {
    int cd;
    memset(v, 0, sizeof(v));
    scanf("%d", &n);
    for (int i = 1; i <= 2 * n; i += 2) {
      scanf("%d", &a[i]);
      v[a[i]] = 1;
    }
    int num = 0;
    for (int i = 2; i <= 2 * n; i += 2) {
      if (a[i - 1] == 2 * n) {
        num = 1;
        break;
      }
      for (int j = a[i - 1] + 1; j <= 2 * n; j++) {
        if (v[j] == 0) {
          a[i] = j;
          v[j] = 1;
          break;
        }
        if (j == 2 * n) {
          num = 1;
          break;
        }
      }
      if (num == 1) break;
    }
    if (num == 1)
      printf("-1\n");
    else {
      for (int i = 1; i <= 2 * n; i++) {
        printf("%d ", a[i]);
      }
      puts("");
    }
  }
}