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>
const long long MX = 1e18, N = 1e4 + 10;
const double PI = 3.141592653589793238;
using namespace std;
long long n, m, ans[N], a[N], t;
map<int, int> mp;
int main() {
  cin >> t;
  while (t--) {
    cin >> n;
    bool yes = true;
    for (int i = 1; i <= n; ++i) {
      cin >> a[i];
      ++mp[a[i]];
    }
    for (int i = 1; i <= n; ++i) {
      bool ok = false;
      ans[i * 2 - 1] = a[i];
      for (int j = a[i] + 1; j <= n * 2; ++j) {
        if (!mp[j]) {
          ans[i * 2] = j;
          mp[j]++;
          ok = true;
          break;
        }
      }
      if (!ok) yes = false;
    }
    if (yes) {
      for (int i = 1; i <= n * 2; i++) cout << ans[i] << ' ';
    } else
      cout << -1;
    cout << '\n';
    mp.clear();
  }
  return 0;
}