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;
long long M = 1000000007;
int main() {
  ios_base::sync_with_stdio(false);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<int> v;
    map<int, int> m;
    int z = 0;
    int j;
    int b[n];
    for (long long i = (0); i < (n); i++) {
      cin >> b[i];
      m[b[i]]++;
    }
    for (long long i = (0); i < (n); i++) {
      j = 1;
      v.push_back(b[i]);
      while (m[b[i] + j] != 0) {
        j++;
      }
      v.push_back(b[i] + j);
      m[b[i] + j]++;
    }
    vector<int> a(2 * n, 0);
    for (long long i = (0); i < (2 * n); i++) {
      a[i] = v[i];
    }
    sort(v.begin(), v.end());
    for (long long i = (0); i < (2 * n); i++) {
      if (v[i] != i + 1) {
        z++;
      }
    }
    if (z == 0) {
      for (long long i = (0); i < (2 * n); i++) {
        cout << a[i] << " ";
      }
      cout << endl;
    } else {
      cout << -1 << endl;
    }
  }
}