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 int dif(long long int a, long long int b) {
  return ((a / b) + (a % b != 0));
}
vector<long long int> v;
long long int c[100100], par[100100];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int t;
  cin >> t;
  while (t--) {
    long long int n, f = 0, k;
    cin >> n;
    pair<long long int, long long int> a[n], sol[2 * n];
    v.clear();
    for (long long int i = 0; i < 2 * n + 2; i++) {
      par[i] = 0;
      c[i] = 0;
    }
    for (long long int i = 0; i < n; i++) {
      cin >> a[i].first;
      a[i].second = i;
      c[a[i].first] = 1;
      if (a[i].first >= 2 * n) {
        f = 1;
      }
    }
    if (f) {
      cout << -1 << '\n';
      continue;
    }
    sort(a, a + n);
    for (long long int i = 1; i <= 2 * n; i++) {
      if (!c[i]) v.push_back(i);
    }
    for (long long int i = 0; i < n; i++) {
      k = lower_bound(v.begin(), v.end(), a[i].first) - v.begin();
      if (k == v.size()) {
        f = 1;
        break;
      }
      par[a[i].first] = v[k];
      v.erase(v.begin() + k);
    }
    if (f) {
      cout << -1 << '\n';
      continue;
    }
    for (long long int i = 0; i < n; i++) {
      sol[a[i].second] = {a[i].first, par[a[i].first]};
    }
    for (long long int i = 0; i < n; i++) {
      for (long long int j = i + 1; j < n; j++) {
        if (sol[i].second > sol[j].second && sol[i].first < sol[j].second &&
            sol[j].first < sol[i].second) {
          swap(sol[j].second, sol[i].second);
        }
      }
    }
    for (long long int i = 0; i < n; i++) {
      cout << min(sol[i].first, sol[i].second) << " "
           << max(sol[i].first, sol[i].second) << " ";
    }
    cout << '\n';
  }
  return 0;
}