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 cinn() {
  int x;
  scanf("%d", &x);
  return x;
}
long long cinl() {
  long long x;
  scanf("%lld", &x);
  return x;
}
void coutt(int x) { printf("%d ", x); }
vector<string> split(const string &s, char c) {
  vector<string> v;
  stringstream ss(s);
  string x;
  while (getline(ss, x, c)) v.emplace_back(x);
  return move(v);
}
void err(vector<string>::iterator it) {}
template <typename T, typename... Args>
void err(vector<string>::iterator it, T a, Args... args) {
  cout << it->substr((*it)[0] == ' ', it->length()) << " = " << a << " ";
  err(++it, args...);
}
const int mod = 1e9 + 7;
const int inf = (int)2e9 + 5;
const long long int Inf = (long long int)1e18 + 5;
const int N = 1e7 + 5;
const int nn = 3e5 + 5;
int siv[2000];
int solve() {
  int n;
  cin >> n;
  int a[106];
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    siv[a[i]] = 1;
  }
  bool f = true;
  std::vector<int> v;
  for (int i = 0; i < n; i++) {
    for (int j = 1; j <= 2 * n; j++) {
      if (a[i] < j && siv[j] == 0) {
        siv[j] = 1;
        v.push_back(a[i]);
        v.push_back(j);
        break;
      }
    }
  }
  for (int i = 1; i <= 2 * n; i++) {
    if (siv[i] == 0) {
      f = false;
    }
  }
  if (f == false) {
    cout << -1 << endl;
  } else {
    for (auto i : v) {
      cout << i << " ";
    }
    printf("\n");
    ;
  }
  for (int i = 0; i <= 2 * n; i++) {
    siv[i] = 0;
  }
  v.clear();
  f = true;
  return 0;
}
int main() {
  int test = 1, tc = 0;
  scanf("%d", &test);
  while (test--) solve();
  return 0;
}