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<long long> vin(long long n) {
  vector<long long> a(n);
  for (long long i = 0; i < n; i += 1) cin >> a[i];
  return a;
}
long long intin() {
  long long n;
  cin >> n;
  return n;
}
char charin() {
  char a;
  cin >> a;
  return a;
}
string strin() {
  string s;
  cin >> s;
  return s;
}
long long factorial(long long n) {
  return (n == 1 || n == 0) ? 1 : n * factorial(n - 1);
}
void solve() {
  long long n;
  cin >> n;
  vector<long long> a = vin(n);
  set<long long> s;
  for (long long i = 1; i < 2 * n + 1; i += 1) s.insert(i);
  for (auto i : a) {
    if (s.find(i) == s.end()) {
      cout << -1 << endl;
      return;
    }
    s.erase(i);
  }
  vector<long long> b(2 * n, 0);
  for (long long i = 0; i < n; i += 1) {
    b[2 * i] = a[i];
  }
  for (long long i = 1; i < 2 * n; i += 2) {
    auto x = upper_bound(s.begin(), s.end(), b[i - 1]);
    if (x == s.end()) {
      cout << -1 << endl;
      return;
    }
    s.erase(x);
    b[i] = *x;
  }
  for (long long i = 0; i < n; i += 1) {
    if (a[i] != min(b[2 * i], b[2 * i + 1])) {
      cout << -1 << endl;
      return;
    }
  }
  for (auto i : b) cout << i << " ";
  cout << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  for (long long _ = 0; _ < t; _ += 1) {
    solve();
  }
  return 0;
}