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;
vector<int> solve(int N, const vector<int>& B) {
  vector<int> pos(2 * N + 1, -1);
  for (int i = 0, _n = (N); i < _n; ++i) {
    int x = B[i];
    pos[x] = i;
  }
  set<int> missing;
  for (int n = (1), _b = (2 * N); n <= _b; ++n) {
    if (pos[n] < 0) missing.insert(n);
  }
  vector<int> A(2 * N);
  for (int i = 0, _n = (N); i < _n; ++i) {
    int x = B[i];
    auto it = missing.upper_bound(x);
    if (it == missing.end()) return {};
    A[i * 2] = x;
    A[i * 2 + 1] = *it;
    missing.erase(it);
  }
  return A;
}
int main(int argc, char* argv[]) {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int TC;
  cin >> TC;
  for (int tc = (1), _b = (TC); tc <= _b; ++tc) {
    int N;
    cin >> N;
    vector<int> B(N);
    for (int i = 0, _n = (N); i < _n; ++i) cin >> B[i];
    vector<int> A = solve(N, B);
    if (A.empty())
      cout << "-1\n";
    else {
      for (int x : A) cout << x << ' ';
      cout << '\n';
    }
  }
  return 0;
}