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 int fun(vector<long long int> &v, long long int t, long long int l,
                  long long int r) {
  if (l == r) return l;
  if (abs(l - r) == 1) {
    long long int t1 = abs(5 * v[l] - 2 * t);
    long long int t2 = abs(5 * v[r] * 2 * t);
    if (t1 < t2)
      return l;
    else
      return r;
  }
  long long int m = (l + r) / 2;
  long long int t1 = abs(5 * v[m - 1] - 2 * t);
  long long int t2 = abs(5 * v[m + 1] - 2 * t);
  if (t1 < t2)
    return fun(v, t, l, m);
  else
    return fun(v, t, m, r);
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout << setprecision(10) << fixed;
  long long int t;
  cin >> t;
  while (t--) {
    long long int n;
    cin >> n;
    long long int b[n];
    map<long long int, long long int> m;
    for (long long int i = 0; i < n; i++) {
      cin >> b[i];
      m[b[i]] = 1;
    }
    long long int f = 0;
    vector<long long int> v;
    for (long long int i = 0; i < n; i++) {
      long long int t1 = b[i] + 1;
      while (m[t1] != 0) t1++;
      if (t1 > n * 2) {
        f = 1;
        break;
      } else {
        v.push_back(b[i]);
        v.push_back(t1);
        m[t1] = 1;
      }
    }
    if (f == 1)
      cout << "-1" << endl;
    else {
      for (long long int i = 0; i < v.size(); i++) cout << v[i] << " ";
      cout << endl;
    }
  }
  return 0;
}