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 t, n;
int main() {
  cin >> t;
  while (t--) {
    int f = 0;
    cin >> n;
    vector<int> a;
    a.push_back(0);
    map<int, int> m;
    int b[2 * n + 5];
    for (int i = 1; i <= n; i++) {
      int x;
      cin >> x;
      if (x >= 2 * n) f = 1;
      a.push_back(x);
      m[x]++;
    }
    if (f) {
      cout << -1 << endl;
      continue;
    }
    for (int i = 1; i <= n && !f; i++) {
      int x = a[i];
      while (m.find(x) != m.end()) x++;
      if (x > 2 * n) {
        f = 1;
      } else {
        m[x]++;
        b[2 * i - 1] = a[i];
        b[2 * i] = x;
      }
    }
    if (f)
      cout << "-1" << endl;
    else {
      for (int i = 1; i <= 2 * n; i++) {
        cout << b[i] << " ";
      }
      cout << endl;
    }
  }
  return 0;
}