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;
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<bool> used(2 * n + 1, 0);
    vector<int> ind(2 * n + 1, -1);
    vector<bool> ind_used(2 * n + 1, 0);
    for (int i = 0; i < n; i++) {
      int b;
      cin >> b;
      used[b] = 1;
      ind[b] = (i + 1) * 2;
      ind_used[(i + 1) * 2] = 1;
    }
    bool possible = true;
    for (int i = 2 * n; i >= 1; i--) {
      if (used[i]) {
        continue;
      }
      int idx = -1;
      for (int j = i - 1; j >= 1; j--) {
        if (ind[j] % 2 == 0 && !ind_used[ind[j] - 1]) {
          idx = ind[j] - 1;
          break;
        }
      }
      if (idx == -1) {
        possible = false;
        break;
      } else {
        ind[i] = idx;
        ind_used[idx] = 1;
      }
    }
    if (!possible) {
      cout << "-1\n";
    } else {
      vector<int> ans(2 * n);
      for (int i = 0; i < 2 * n; i++) {
        ans[ind[i + 1] - 1] = i + 1;
      }
      for (int i = 0; i < 2 * n; i += 2) {
        for (int j = i + 2; j < 2 * n; j += 2) {
          if (ans[i] > ans[j] && ans[i + 1] < ans[j] && ans[j + 1] < ans[i]) {
            swap(ans[i], ans[j]);
          }
        }
      }
      for (int i = 0; i < 2 * n; i += 2) {
        if (ans[i] > ans[i + 1]) {
          swap(ans[i], ans[i + 1]);
        }
      }
      for (int i = 0; i < 2 * n; i++) {
        cout << ans[i] << " ";
      }
      cout << endl;
    }
  }
  return 0;
}