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, n, i, j, k, a, b, c, m, l, r, s;
void dihan() {
  cin >> n;
  vector<long long> x(n);
  for (auto &i : x) cin >> i;
  vector<long long> y(2 * n + 1);
  for (i = 0; i < n; i++) y[x[i]] = i + 1;
  priority_queue<long long> q;
  vector<pair<long long, long long> > p;
  for (i = 1; i < 2 * n + 1; i++)
    if (y[i] != 0)
      q.push(-y[i]);
    else if (q.size() > 0) {
      y[i] = -q.top();
      q.pop();
    } else {
      cout << -1 << endl;
      return;
    }
  for (i = 1; i < 2 * n + 1; i++) p.push_back({y[i], i});
  sort(p.begin(), p.end());
  for (auto i : p) cout << i.second << " ";
  cout << endl;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(nullptr);
  cin >> t;
  while (t--) dihan();
}