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;
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    pair<long long, long long> a[n];
    long long re[2 * n + 1];
    map<long long, long long> mp;
    for (long long i = 0; i < n; i++) {
      cin >> a[i].first;
      mp[a[i].first] = 1;
      a[i].second = i;
    }
    bool ch = true;
    long long ma = 2 * n;
    for (long long i = 0; i < n; i++) {
      ma = a[i].first;
      while (ma <= 2 * n && mp[ma] == 1) {
        ma++;
      }
      if (ma > 2 * n) {
        ch = false;
        break;
      }
      if (a[i].second % 2 == 0) {
        re[2 * a[i].second] = a[i].first;
        re[2 * a[i].second + 1] = ma;
        mp[ma] = 1;
      } else {
        re[2 * a[i].second] = a[i].first;
        re[2 * a[i].second + 1] = ma;
        mp[ma] = 1;
      }
    }
    if (!ch)
      cout << -1 << endl;
    else {
      for (long long i = 0; i < 2 * n; i++) cout << re[i] << " ";
      cout << endl;
    }
  }
  return 0;
}