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 b[105], vis[205];
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    int n;
    scanf("%d", &n);
    int pd = 0;
    memset(vis, 0, sizeof(vis));
    int maxx = 0;
    for (int i = 0; i < n; i++) {
      scanf("%d", &b[i]);
      maxx = max(maxx, b[i]);
      vis[b[i]] = 1;
    }
    if (maxx >= 2 * n) pd = 1;
    if (pd) {
      printf("-1\n");
      continue;
    }
    vector<int> e;
    int pos = 1;
    for (int i = 0; i < n; i++) {
      pos = 1;
      while (pos <= b[i] || vis[pos] == 1) pos++;
      if (pos > 2 * n) break;
      vis[pos] = 1;
      e.push_back(pos);
    }
    if (e.size() != n) {
      printf("-1\n");
      continue;
    }
    for (int i = 0; i < n; i++) printf("%d %d ", b[i], e[i]);
    printf("\n");
  }
}