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() {
  int T;
  cin >> T;
  bool used[314] = {};
  int A[210] = {};
  while (T--) {
    int N;
    cin >> N;
    for (int i = 0; i < N; i++) {
      cin >> A[i * 2];
      used[A[i * 2]] = true;
    }
    for (int i = 0; i < N; i++) {
      int j = A[i * 2] + 1;
      while (used[j]) j++;
      A[i * 2 + 1] = j;
      used[j] = true;
    }
    bool ok = true;
    for (int i = 1; i <= N * 2; i++) {
      if (!used[i]) ok = false;
    }
    if (!ok)
      cout << -1 << endl;
    else {
      for (int i = 0; i < N * 2; i++) cout << A[i] << " ";
      cout << endl;
    }
    ok = true;
    fill(used, used + 310, false);
  }
}