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 t, n;
void solve() {
  cin >> n;
  int b[n];
  bool foundMin = false;
  bool foundMax = false;
  for (int i = 0; i < n; i++) {
    cin >> b[i];
    if (b[i] == 1) foundMin = true;
    if (b[i] == n * 2) foundMax = true;
  }
  if (!foundMin || foundMax) {
    cout << "-1\n";
    return;
  }
  int tempB[n];
  for (int i = 0; i < n; i++) {
    tempB[i] = b[i];
  }
  sort(tempB, tempB + n);
  int lastNum = 1;
  int needed = 2;
  for (int i = 1; i < n; i++) {
    if (tempB[i] - lastNum > needed) {
      cout << "-1\n";
      return;
    } else {
      needed += tempB[i] - lastNum;
      lastNum = tempB[i];
    }
  }
  bool taken[n * 2];
  for (int i = 0; i < n * 2; i++) {
    taken[i] = false;
  }
  for (int i = 0; i < n; i++) {
    taken[b[i] - 1] = true;
  }
  vector<int> ans;
  for (int i = 0; i < n; i++) {
    ans.push_back(b[i]);
    bool found = false;
    for (int j = b[i]; j < n * 2; j++) {
      if (!taken[j]) {
        ans.push_back(j + 1);
        taken[j] = true;
        found = true;
        break;
      }
    }
    if (!found) {
      cout << "-1\n";
      return;
    }
  }
  for (int i = 0; i < ans.size(); i++) cout << ans[i] << " ";
  cout << "\n";
}
int main(int argc, char **argv) {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> t;
  for (int i = 0; i < t; i++) {
    solve();
  }
  return 0;
}