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 int N = int(2e5 + 20);
const int INF = int(1e9 + 5);
const long long LINF = int(1e18 + 20);
int main() {
  int t;
  cin >> t;
  for (; t; t--) {
    int n;
    cin >> n;
    set<int> ok;
    for (int i = 1; i <= 2 * n; i++) ok.insert(i);
    vector<int> a(n);
    for (int i = 0; i < n; i++) {
      cin >> a[i];
      ok.erase(a[i]);
    }
    vector<int> b(2 * n);
    for (int i = 0; i < 2 * n; i += 2) {
      b[i] = a[i / 2];
    }
    bool f = 0;
    for (int i = 1; i < 2 * n; i += 2) {
      if (ok.lower_bound(b[i - 1]) == ok.end()) {
        f = 1;
        break;
      }
      b[i] = *ok.lower_bound(b[i - 1]);
      ok.erase(ok.lower_bound(b[i - 1]));
    }
    if (f) {
      cout << "-1\n";
      continue;
    }
    for (int i = 1; i < 2 * n; i += 2)
      for (int j = i + 2; j < 2 * n; j += 2) {
        if (b[i] > b[j] && b[i] > b[j - 1] && b[j] > b[i - 1]) swap(b[i], b[j]);
      }
    for (auto i : b) cout << i << ' ';
    cout << '\n';
  }
}