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 int N = 1e5 + 55;
int a[N];
pair<int, int> c[N];
int b[N];
bool vis[N];
int main() {
  int t;
  cin >> t;
  while (t--) {
    memset(vis, 0, sizeof vis);
    int n;
    cin >> n;
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
      vis[a[i]] = 1;
      c[i] = {a[i], i};
      b[i * 2 - 1] = a[i];
    }
    bool q = 0;
    for (int i = 1; i <= n; i++) {
      int j;
      for (j = a[i] + 1; j <= n * 2; j++) {
        if (vis[j] == 0) {
          vis[j] = 1;
          b[i * 2] = j;
          break;
        }
      }
      if (j == n * 2 + 1) {
        q = 1;
        cout << "-1\n";
        break;
      }
    }
    if (!q)
      for (int i = 1; i <= n; i++) {
        if (a[i] != min(b[i * 2], b[i * 2 - 1])) {
          q = 1;
          cout << "-1\n";
          break;
        }
      }
    if (!q) {
      for (int i = 1; i <= n * 2; i++) cout << b[i] << ' ';
      cout << endl;
    }
  }
  return 0;
}