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 t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    long long b[n];
    for (long long i = 0; i < n; i++) cin >> b[i];
    bool c[2 * n + 1];
    memset(c, true, 2 * n + 1);
    long long a[2 * n];
    memset(a, -1, 2 * n);
    for (long long i = 0; i < n; i++) {
      a[2 * (i + 1) - 1] = b[i];
      c[b[i]] = false;
    }
    bool T = true;
    for (long long i = 0; i < 2 * n; i += 2) {
      long long j = a[i + 1];
      while (c[j] == false && j <= 2 * n) {
        j++;
      }
      if (j > 2 * n) {
        T = false;
        break;
      }
      a[i] = j;
      c[j] = false;
    }
    if (!T) {
      cout << -1 << endl;
      continue;
    }
    for (long long i = 0; i < 2 * n; i += 2) {
      if (a[i] > a[i + 1]) {
        swap(a[i], a[i + 1]);
      }
    }
    for (long long i = 0; i < 2 * n; i++) {
      cout << a[i] << " ";
    }
    cout << endl;
  }
}