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;
const long long mod = 1e9 + 7;
const long long inf = 0x3f3f3f3f;
const long long nax = 0;
void solve() {
  long long n;
  vector<long long> b(210, 0), arr(105);
  vector<bool> vis(210, 0);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> arr[i];
    vis[arr[i]] = 1;
  }
  for (int i = 1; i <= n; ++i) {
    b[2 * i - 1] = arr[i];
    int angka = arr[i] + 1;
    while (vis[angka] && angka <= 2 * n) {
      if (angka == 2 * n) {
        angka = -1;
        break;
      }
      angka++;
    }
    if (angka == -1 || angka > n * 2) {
      cout << "-1" << '\n';
      return;
    }
    vis[angka] = 1;
    b[2 * i] = angka;
  }
  for (int i = 1; i <= 2 * n; ++i) {
    cout << b[i] << ' ';
  }
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie();
  cout.tie();
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
}