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 f(map<int, int> &mp, int b[], int n, int l) {
  bool x = 0;
  for (int j = l + 1; j <= 2 * n; j++) {
    if (!mp[j]) {
      mp[j] = 1;
      x = 1;
      return j;
    }
  }
  return 0;
}
int main() {
  int T;
  cin >> T;
  while (T--) {
    int n;
    cin >> n;
    int a[n + 3], b[2 * n + 5];
    map<int, int> mp;
    for (int i = 1; i <= n; i++) cin >> a[i], mp[a[i]]++;
    bool y = 0;
    int x = 1;
    for (int i = 1; i <= 2 * n; i += 2, x++) {
      int k = f(mp, b, n, a[x]);
      if (!k) {
        y = 1;
        break;
      }
      b[i + 1] = k;
      b[i] = a[x];
    }
    if (y) {
      cout << "-1\n";
      continue;
    }
    x = 1;
    for (int i = 1; i <= 2 * n; i += 2, x++) {
      if (a[x] == min(b[i], b[i + 1])) continue;
      cout << "-1\n";
      y = 1;
      break;
    }
    if (!y) {
      for (int i = 1; i <= 2 * n; i++) cout << b[i] << " \n"[i == 2 * n];
    }
  }
}