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 arr[106];
int memo[206];
int solve(int n, int reff) {
  for (int i = 1; i <= n; i++) {
    if ((memo[i] == -1) && i > reff) {
      memo[i] = 1;
      return i;
    }
  }
  return 0;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n, x;
    memset(memo, -1, sizeof memo);
    cin >> n;
    int cnt = 1;
    for (int i = 0; i < n; i++) {
      cin >> x;
      arr[cnt] = x;
      memo[x] = 1;
      cnt += 2;
    }
    n *= 2;
    bool chk = false;
    for (int i = 1; i <= n; i++) {
      if (!(i & 1)) {
        arr[i] = solve(n, arr[i - 1]);
      }
      if (!arr[i]) {
        chk = true;
        break;
      }
    }
    if (chk) {
      cout << -1 << endl;
      continue;
    }
    for (int i = 1; i <= n; i++) cout << arr[i] << ' ';
    cout << endl;
  }
  return 0;
}