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 a[101];
bool c[201];
vector<int> b[101];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    memset(c, false, sizeof(c));
    for (int i = 0; i < n; i++) {
      b[i].clear();
      cin >> a[i];
      c[a[i]] = true;
      b[i].push_back(a[i]);
    }
    bool ff = false;
    for (int i = 1; i <= 2 * n; i++) {
      if (!c[i]) {
        bool f = false;
        for (int j = 0; j < n; j++) {
          if (b[j].size() == 2) continue;
          if (b[j][0] < i) {
            b[j].push_back(i);
            sort(b[j].begin(), b[j].end());
            f = true;
            break;
          }
        }
        if (!f) {
          ff = true;
          break;
        }
      }
    }
    if (ff) {
      cout << -1 << '\n';
    } else {
      for (int i = 0; i < n; i++) {
        cout << b[i][0] << ' ' << b[i][1] << ' ';
      }
      cout << '\n';
    }
  }
}