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;
void compute() {
  int n;
  cin >> n;
  vector<int> ft(n);
  vector<int> res(2 * n);
  for (int i = 0; i < n; ++i) {
    cin >> ft[i];
    ft[i]--;
  }
  set<int> st;
  for (int i = 0; i < 2 * n; ++i) {
    st.insert(i);
  }
  for (int i = 0; i < n; ++i) {
    st.erase(st.find(ft[i]));
  }
  for (int i = 0; i < n; ++i) {
    res[2 * i] = ft[i];
    set<int>::iterator it = st.upper_bound(ft[i]);
    if (it == st.end()) {
      printf("%d\n", -1);
      return;
    }
    res[2 * i + 1] = *it;
    st.erase(it);
  }
  for (int i = 0; i < 2 * n; ++i) {
    printf("%d ", res[i] + 1);
  }
  printf("\n");
}
int main() {
  int t;
  cin >> t;
  for (int tc = 0; tc < t; ++tc) {
    compute();
  }
  return 0;
}