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 = 100 + 3;
int a[N], b[N], vis[300];
int main() {
  int t, n, x;
  map<int, int> mp;
  cin >> t;
  while (t--) {
    cin >> n;
    int jg = 0, bj = 0;
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
      b[i * 2 - 1] = a[i];
      vis[a[i]]++;
      if (a[i] >= n * 2) bj = 1;
    }
    if (bj == 1)
      printf("-1\n");
    else {
      for (int i = 2; i <= n * 2; i += 2) {
        int ks = b[i - 1];
        while (vis[ks] == 1) ks++;
        b[i] = ks;
        vis[ks]++;
        if (ks > n * 2) {
          jg = 1;
          break;
        }
      }
      if (jg == 1)
        printf("-1\n");
      else {
        for (int i = 1; i <= 2 * n; i++) {
          printf("%d ", b[i]);
        }
        printf("\n");
      }
    }
    for (int i = 1; i <= 200; i++) vis[i] = 0;
  }
  return 0;
}