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 main() {
  bool u = 0, n = 0, z = 0;
  int a, b, c[100000] = {}, p, d, q = 1, s = 1;
  cin >> a;
  for (int i = 0; i < a; i++) {
    cin >> b;
    p = b * 2;
    s = 1;
    for (int i = 0; i <= (b * 2 - 2); i += 2) cin >> c[i];
    for (int i = 0; i <= (b * 2 - 2); i += 2) {
      u = 0;
      q = 1;
      while (1) {
        d = c[i] + q;
        if (d > p) {
          u = 1;
          break;
        } else if (p >= d) {
          n = 0;
          z = 0;
          for (int i = 0; i <= b * 2; i++)
            if (d == c[i]) {
              n = 1;
              break;
            }
          if (n == 0) {
            c[s] = d;
            s += 2;
            z = 1;
            break;
          }
          if (n == 1 && i == b * 2 - 1) {
            u = 1;
            break;
          }
        }
        q++;
      }
      if (u == 1) break;
    }
    if (u == 1)
      cout << -1 << endl;
    else {
      for (int i = 0; i < b * 2; i++) cout << c[i] << " ";
    }
    cout << endl;
    for (int i = 0; i < 10000; i++) c[i] = 0;
  }
  return 0;
}