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;
void solve() {
  long long int t, m, i, j, k, sum = 0;
  long long int n;
  cin >> n;
  vector<pair<long long int, long long int>> a(n);
  vector<long long int> viz(2 * n + 5, 0);
  for (i = 0; i < n; i++) {
    cin >> a[i].first;
    a[i].second = i;
    viz[a[i].first] = 1;
  }
  sort(a.begin(), a.end(), greater<pair<long long int, long long int>>());
  vector<long long int> ans(2 * n + 1);
  for (i = 0; i < n; i++) {
    long long int flag = 0;
    for (j = 2 * n; j > a[i].first; j--) {
      if (!viz[j]) {
        viz[j] = 1;
        ans[2 * a[i].second] = a[i].first;
        ans[2 * a[i].second + 1] = j;
        flag = 1;
        break;
      }
    }
    if (!flag) {
      cout << -1 << '\n';
      return;
    }
  }
  for (i = 1; i < 2 * n; i += 2) {
    for (j = i + 2; j < 2 * n; j += 2) {
      if (ans[i] > ans[j] && ans[j] > ans[i - 1] && ans[i] > ans[j - 1]) {
        swap(ans[i], ans[j]);
      }
    }
  }
  for (i = 0; i < 2 * n; i++) {
    cout << ans[i] << ' ';
  }
  cout << '\n';
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int q;
  cin >> q;
  while (q--) {
    solve();
  }
  return 0;
}