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;
bool compare(string &s1, string &s2) { return s1.size() < s2.size(); }
string lower(string second) {
  transform(second.begin(), second.end(), second.begin(), ::tolower);
  return second;
}
string upper(string second) {
  transform(second.begin(), second.end(), second.begin(), ::toupper);
  return second;
}
map<int, int> mp;
set<int> myset;
vector<int> v, v1, v2;
long long n, m, i, j, k, h, t, x = 1, y = 1, z = 1, d = 0;
int ans = 0, sum = 0, cnt = 0, p = 0, mx = 0, mn = 0;
int a, b, l, r;
string second, st, str;
int main() {
  cin >> t;
  while (t--) {
    cin >> n;
    for (i = 1; i <= 2 * n; i++) mp[i];
    int a[n + 5];
    for (i = 1; i <= n; i++) {
      cin >> a[i];
      mp[a[i]]++;
    }
    for (i = 1; i <= n; i++) {
      x = a[i];
      v.push_back(x);
      for (auto it : mp) {
        if (it.second == 0 and it.first > x) {
          mp[it.first]++;
          v.push_back(it.first);
          break;
        }
      }
    }
    if (v.size() == 2 * n) {
      for (auto it : v) cout << it << ' ';
    } else {
      cout << -1;
    }
    cout << endl;
    v.clear();
    mp.clear();
  }
}