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 int t, i, j, k, n, x, y, x1, y1, x2, y2, p, arr[1000], b[2000], l;
  cin >> t;
  while (t--) {
    cin >> n;
    j = 0;
    long long int flag = 0, flag1 = 1;
    for (i = 0; i < n; i++) {
      cin >> arr[i];
      if (arr[i] > n) {
        j++;
      }
      if (arr[i] == 2 * n) {
        flag = 1;
      }
      if (arr[i] == 1) {
        flag1 = 0;
      }
    }
    if (j > n / 2 || flag) {
      cout << "-1\n";
    } else {
      set<long long int> se;
      for (i = 1; i <= 2 * n; i++) {
        se.insert(i);
      }
      for (i = 0; i < n; i++) {
        se.erase(arr[i]);
      }
      l = 0;
      long long int flag2 = 0;
      for (i = 0; i < n; i++) {
        b[l++] = arr[i];
        j = arr[i] + 1;
        long long int flag3 = 0;
        for (j = j; j <= 2 * n; j++) {
          if (se.find(j) != se.end()) {
            b[l++] = j;
            se.erase(j);
            flag3 = 1;
            break;
          }
        }
        if (flag3 == 0) {
          flag2 = 1;
          break;
        }
      }
      if (flag2) {
        cout << "-1\n";
        continue;
      }
      for (i = 0; i < 2 * n; i++) {
        cout << b[i] << " ";
      }
      cout << endl;
    }
  }
  return 0;
}