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 gcd(long long a, long long b) {
  if (a == 0)
    return b;
  else
    return gcd(b % a, a);
}
long long findGCD(vector<long long> arr, long long n) {
  long long result = arr[0];
  for (long long i = 1; i < n; i++) result = gcd(arr[i], result);
  return result;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout.precision(8);
  cout << fixed;
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    bool flag = true;
    vector<long long> v(n);
    for (auto &itr : v) cin >> itr;
    long long maxx = 2 * n;
    vector<long long> ans;
    for (auto ele : v) {
      long long nxt = ele + 1;
      while (1) {
        if (nxt > maxx) {
          flag = false;
          break;
        }
        auto pos = find(v.begin(), v.end(), nxt);
        auto pos2 = find(ans.begin(), ans.end(), nxt);
        if (pos == v.end() && pos2 == ans.end()) {
          ans.push_back(ele);
          ans.push_back(nxt);
          break;
        } else {
          nxt++;
        }
      }
      if (!flag) break;
    }
    if (!flag)
      cout << "-1";
    else
      for (auto ele : ans) cout << ele << " ";
    cout << endl;
  }
  return 0;
}