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;
const long long mod = 1e9 + 7;
long long power(long long a, long long b, long long m = mod) {
  long long x = 1;
  while (b) {
    if (b & 1) {
      x = 1ll * x * a % m;
    }
    a = 1ll * a * a % m;
    b /= 2;
  }
  return x;
}
const int N = 1e5 + 9;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int tt;
  cin >> tt;
  while (tt--) {
    int n;
    cin >> n;
    vector<int> a(n + 1), taken(2 * n + 1), b(2 * n + 1);
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
      b[2 * i - 1] = a[i];
      taken[a[i]] = 1;
    }
    set<int> ss;
    for (int i = 1; i <= 2 * n; i++) {
      if (taken[i] == 0) ss.insert(i);
    }
    bool can = 1;
    for (int i = 0; i < n; i++) {
      auto it = ss.upper_bound(b[2 * i + 1]);
      if (it == ss.end()) {
        can = 0;
        break;
      }
      b[2 * i + 2] = *it;
      ss.erase(it);
    }
    if (can) {
      for (int i = 1; i <= 2 * n; i++) {
        cout << b[i] << " ";
      }
      cout << "\n";
    } else {
      cout << "-1\n";
    }
  }
  return 0;
}