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;
long long t, n, b[100009], f[100009], j, fl;
struct st {
  long long x, i;
} a[100009];
bool cmp(st a, st b) { return a.x < b.x; }
bool cm(st a, st b) { return a.i < b.i; }
int main() {
  cin >> t;
  while (t--) {
    j = 1;
    fl = 0;
    cin >> n;
    for (int i = 0; i <= n * 2 + 50; i++) b[i] = 0, f[i] = 0;
    for (int i = 1; i <= n; i++) {
      cin >> a[i].x;
      a[i].i = i;
      f[a[i].x] = 1;
    }
    for (int i = 1; i <= n; i++) {
      j = a[i].x;
      while (f[j]) j++;
      b[a[i].i * 2 - 1] = a[i].x;
      b[a[i].i * 2] = j;
      f[j] = 1;
    }
    for (int i = 1; i <= n; i++) {
      if (a[i].x != min(b[i * 2], b[i * 2 - 1]) || b[i * 2] > 2 * n ||
          b[i * 2 - 1] > 2 * n) {
        fl = 1;
        break;
      }
    }
    if (fl)
      cout << -1 << endl;
    else {
      for (int i = 1; i <= n * 2; i++) cout << b[i] << " ";
      cout << endl;
    }
  }
}