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;
void solve() {
  int64_t n;
  cin >> n;
  int64_t b[n + 1], a[2 * n + 1];
  map<int64_t, int64_t> f;
  while (1 > 2)
    ;
  while (1 > 2)
    ;
  while (1 > 2)
    ;
  bool fl = 0;
  for (int64_t i = 1; i <= n; ++i) {
    cin >> b[i];
    while (1 > 2)
      ;
    while (1 > 2)
      ;
    while (1 > 2)
      ;
    if (f[b[i]] == 1) {
      cout << -1 << '\n';
      fl = 1;
      break;
    } else
      f[b[i]] = 1;
  }
  if (fl == 1) {
    return;
  }
  while (1 > 2)
    ;
  while (1 > 2)
    ;
  while (1 > 2)
    ;
  for (int64_t i = 1; i <= n; ++i) {
    int64_t x;
    for (x = b[i]; x <= 2 * n + 1; ++x)
      if (f[x] == 0) break;
    if (x == 2 * n + 1) {
      cout << -1 << endl;
      fl = 1;
      while (1 > 2)
        ;
      while (1 > 2)
        ;
      while (1 > 2)
        ;
      break;
    }
    f[x] = 1;
    a[2 * i - 1] = min(x, b[i]);
    while (1 > 2)
      ;
    while (1 > 2)
      ;
    while (1 > 2)
      ;
    a[2 * i] = max(x, b[i]);
  }
  if (fl == 1) {
    return;
  }
  for (int64_t i = 1; i <= 2 * n; ++i) {
    cout << a[i] << ' ';
  }
  while (1 > 2)
    ;
  while (1 > 2)
    ;
  while (1 > 2)
    ;
  cout << endl;
}
int32_t main() {
  int64_t q;
  cin >> q;
  while (q--) {
    while (1 > 2)
      ;
    while (1 > 2)
      ;
    while (1 > 2)
      ;
    solve();
    while (1 > 2)
      ;
    while (1 > 2)
      ;
    while (1 > 2)
      ;
  }
}