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 main() {
  long long int b, c = 0, sum = 0, i = 0, s, j, n, d = 0, q = 0, e, f, t = 0, x,
                   h, l = 0, k = 0, x1, x2, y1, y2, m;
  cin >> t;
  vector<long long int> v;
  vector<long long int> a;
  vector<long long int> z;
  map<long long int, long long int> mp;
  for (j = 0; j < t; j++) {
    cin >> n;
    v.clear();
    a.clear();
    z.clear();
    mp.clear();
    for (i = 0; i < n; i++) {
      cin >> x;
      v.push_back(x);
      z.push_back(x);
    }
    sort(v.begin(), v.end());
    c = 0;
    s = 1;
    for (k = c + 1; k < 2 * n + 1; k++) {
      for (i = 0; i < n; i++) {
        l = 1;
        if (v[i] == k) {
          l = 0;
          break;
        }
      }
      if (l != 0) {
        c = k;
        a.push_back(c);
      }
    }
    if (v[0] != 1) {
      cout << "-1" << endl;
      continue;
    }
    for (f = 0; f < n; f++) {
      if (v[f] > a[f]) {
        cout << "-1" << endl;
        s = 0;
        break;
      }
    }
    if (s != 0) {
      for (f = 0; f < n; f++) {
        for (i = 0; i < a.size(); i++) {
          if (a[i] > z[f]) {
            cout << z[f] << " " << a[i] << " ";
            a.erase(a.begin() + i);
            break;
          }
        }
      }
    }
    cout << endl;
  }
}