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;
const int N = 2e5 + 10;
const int mod = 1e9 + 7;
void solve() {
  int n, m, i, j, k, l, r, x, y, z, a, b, c;
  cin >> n;
  vector<int> v(n);
  for (int zz = 0; zz < n; zz++) cin >> v[zz];
  bool chk[n * 2 + 5];
  memset(chk, 0, sizeof(chk));
  for (i = 0; i < n; i++) {
    chk[v[i]] = true;
  }
  bool yes;
  vector<int> ans;
  for (i = 0; i < n; i++) {
    ans.push_back(v[i]);
    yes = false;
    for (j = v[i]; j <= 2 * n; j++) {
      if (chk[j] == false) {
        ans.push_back(j);
        chk[j] = true;
        yes = true;
        break;
      }
    }
    if (yes == false) {
      cout << "-1"
           << "\n";
      return;
    }
  }
  for (int zzz = 0; zzz < ans.size(); zzz++) cout << ans[zzz] << " ";
  cout << "\n";
}
int main() {
  int t;
  cin >> t;
  while (t--) solve();
}