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 mark[2005];
int main() {
  long long n, m, i, j, t, x, y, w, k, a, b;
  cin >> t;
  while (t--) {
    cin >> n;
    vector<pair<int, int> > v;
    memset(mark, 0, sizeof mark);
    for (i = 0; i < n; i++) {
      cin >> x;
      v.push_back({x, -1});
      mark[x] = 1;
    }
    set<int> s;
    for (i = 1; i <= 2 * n; i++) {
      if (mark[i] == 0) {
        s.insert(i);
      }
    }
    int flag = 1;
    for (i = 0; i < n; i++) {
      x = v[i].first;
      int ans = -1;
      for (auto xx : s) {
        if (xx > x) {
          ans = xx;
          break;
        }
      }
      if (ans == -1) {
        flag = -1;
        break;
      } else {
        v[i].second = ans;
        s.erase(ans);
      }
    }
    if (flag == -1) {
      cout << (-1) << "\n";
    } else {
      for (i = 0; i < n; i++) {
        cout << min(v[i].first, v[i].second) << " "
             << max(v[i].first, v[i].second) << " ";
      }
      cout << "\n";
    }
  }
  return 0;
}