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 mpow(int base, int exp);
void solve() {
  long long int n;
  cin >> n;
  vector<long long int> v(n);
  vector<long long int> cnt(2 * n + 1, 0);
  for (long long int i = 0; i < (n); i++) {
    cin >> v[i];
    cnt[v[i]]++;
  }
  vector<long long int> out;
  for (long long int i = 0; i < (n); i++) {
    out.push_back(v[i]);
    bool flag = true;
    for (long long int j = v[i] + 1; j <= 2 * n; j++) {
      if (cnt[j] == 0) {
        out.push_back(j);
        cnt[j]++;
        flag = false;
        break;
      }
    }
    if (flag) {
      cout << -1 << '\n';
      return;
    }
  }
  for (auto i : out) cout << i << " ";
  cout << '\n';
  ;
}
int main() {
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}
int mpow(int base, int exp) {
  base %= 1000000007;
  int result = 1;
  while (exp > 0) {
    if (exp & 1) result = ((long long int)result * base) % 1000000007;
    base = ((long long int)base * base) % 1000000007;
    exp >>= 1;
  }
  return result;
}