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;
const long long mod = 1e9 + 7;
const long long inf = 1e18;
const long long dx[8] = {-1, 0, 1, 0, -1, 1, 1, -1},
                dy[8] = {0, 1, 0, -1, 1, 1, -1, -1};
const long long kdx[8] = {-2, -1, 1, 2, 2, 1, -1, -2},
                kdy[8] = {1, 2, 2, 1, -1, -2, -2, -1};
void solve() {
  long long n;
  cin >> n;
  long long f = 0, f2 = 0;
  vector<long long> v(n);
  set<long long> s;
  vector<long long> nots;
  for (long long i = 0; i <= n - 1; ++i) {
    cin >> v[i];
    if (v[i] == 2 * n) {
      f = 1;
    }
    if (v[i] == 1) {
      f2 = 1;
    }
    s.insert(v[i]);
  }
  if (f == 1) {
    cout << "-1\n";
    return;
  }
  if (f2 == 0) {
    cout << "-1\n";
    return;
  }
  for (long long i = 1; i <= 2 * n; ++i)
    if (s.count(i) == 0) nots.push_back(i);
  map<long long, long long> make_pair;
  for (auto x : v) {
    auto idx = upper_bound(nots.begin(), nots.end(), x) - nots.begin();
    if (idx >= (long long)(nots.size())) {
      cout << "-1\n";
      return;
    }
    make_pair[x] = nots[idx];
    nots.erase(nots.begin() + idx);
  }
  for (auto x : v) {
    cout << x << ' ' << make_pair[x] << ' ';
  }
  cout << '\n';
}
int32_t main() {
  clock_t start = clock();
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t = 1;
  cin >> t;
  for (long long i = 1; i <= t; ++i) {
    solve();
  }
  double duration = (clock() - start) / (double)CLOCKS_PER_SEC;
  cerr << duration << 's' << "\n";
  return 0;
}