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;
long long t, i, a[100000], h, b[10000], k[10000], g, p, r, l, n;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> t;
  while (t--) {
    cin >> n;
    for (i = 1; i <= n; i++) {
      cin >> a[i];
      b[a[i]]++;
    }
    i = 0;
    l = 0;
    g = 0;
    h = n;
    while (h-- and l == 0) {
      g++;
      i++;
      k[g] = a[i];
      r = a[i] + 1;
      p = 0;
      while (p == 0) {
        if (b[r] == 0) {
          p++;
          b[r]++;
        }
        r++;
      }
      if (r - 1 > n * 2) {
        l++;
      }
      g++;
      k[g] = r - 1;
    }
    if (l == 1) {
      cout << -1;
    } else {
      for (i = 1; i <= n * 2; i++) {
        cout << k[i] << " ";
      }
    }
    cout << '\n';
    fill(b, b + 3 * n, 0);
  }
}