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 fast_expo(long long int x, long long int p) {
  if (p == 0)
    return 1;
  else if (p % 2 == 0) {
    long long int t = fast_expo(x, p / 2) % 1000000007;
    return (t * t) % 1000000007;
  } else
    return (x * (fast_expo(x, p - 1)) % 1000000007) % 1000000007;
}
int main() {
  int t, i;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    map<int, int> mp;
    vector<int> b(n), a(2 * n);
    for (i = 0; i < n; i++) {
      cin >> b[i];
      mp[b[i]] = 1;
    }
    vector<int> v;
    for (i = 1; i <= 2 * n; i++) {
      if (!mp[i]) v.push_back(i);
    }
    int j = 0;
    for (i = 0; i < 2 * n; i += 2) {
      a[i] = b[j];
      j++;
    }
    int flag = 0;
    sort(v.begin(), v.end());
    for (i = 0; i < 2 * n - 1; i += 2) {
      auto it = lower_bound(v.begin(), v.end(), a[i]) - v.begin();
      auto it1 = lower_bound(v.begin(), v.end(), a[i]);
      if (it1 != v.end()) {
        a[i + 1] = v[it];
        if (v.size() > 0) v.erase(it1);
      } else {
        flag = 1;
        break;
      }
    }
    if (flag)
      cout << -1;
    else {
      for (i = 0; i < 2 * n; i++) cout << a[i] << " ";
    }
    cout << endl;
  }
}