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 double PI = acos(-1.0);
const double EPS = -1e-2;
const int dx[] = {1, 0, -1, 0, 1, 1, -1, -1};
const int dy[] = {0, 1, 0, -1, 1, -1, 1, -1};
const long long mod = 1e9 + 7;
const int Mx = INT_MAX;
const int Mn = INT_MIN;
const long long MX = LLONG_MAX;
const long long MN = LLONG_MIN;
const int N = 1e5 + 9;
void Open() {}
long long arr[100005];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  Open();
  int t;
  cin >> t;
  while (t--) {
    set<int> st;
    set<int>::iterator it;
    ;
    int n;
    cin >> n;
    vector<int> v(n + 1), ans(2 * n + 1);
    for (int i = 1; i <= 2 * n; i++) st.insert(i);
    for (int i = 1; i <= n; i++) {
      cin >> v[i];
      st.erase(v[i]);
    }
    bool ok = 1;
    for (int i = 1; i <= n; i++) {
      it = st.upper_bound(v[i]);
      if (it == st.end()) {
        ok = 0;
        break;
      }
      ans[2 * i - 1] = v[i];
      ans[2 * i] = *it;
      st.erase(it);
    }
    if (ok) {
      for (int i = 1; i <= 2 * n; i++) cout << ans[i] << " ";
    } else
      cout << -1;
    cout << "\n";
  }
}