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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int main() {
  int T;
  cin >> T;
  while (T--) {
    int n;
    cin >> n;
    vector<int> a(n), ans(2 * n), used(2 * n + 5);
    for (int i = 0; i < n; i++) {
      cin >> a[i];
      used[a[i]] = 1;
      ans[i * 2] = a[i];
    }
    for (int i = 1; i < 2 * n; i += 2) {
      for (int j = ans[i - 1] + 1; j <= 2 * n; j++)
        if (!used[j]) {
          ans[i] = j;
          used[j] = 1;
          break;
        }
    }
    bool bad = false;
    for (int i = 0; i < 2 * n; i++)
      if (ans[i] == 0) {
        cout << "-1\n";
        bad = true;
        break;
      }
    if (bad) continue;
    for (int i = 0; i < 2 * n; i++) cout << ans[i] << ' ';
    cout << '\n';
  }
}