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 int MOD = 1000000007;
const int INF = 1000000007;
const int MAXN = (int)3e5 + 1;
void setIO(string name) {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  freopen((name + ".in").c_str(), "r", stdin);
  freopen((name + ".out").c_str(), "w", stdout);
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    bool used[2 * n];
    for (int i = 0; i < (2 * n); i++) {
      used[i] = false;
    }
    int seq[n];
    for (int i = 0; i < n; i++) {
      cin >> seq[i];
      seq[i]--;
      used[seq[i]] = true;
    }
    int ans[2 * n];
    bool can = true;
    for (int i = 0; i < n; i++) {
      int val = seq[i];
      while (val < 2 * n && used[val]) {
        val++;
      }
      if (val == 2 * n) {
        can = false;
        break;
      }
      used[val] = true;
      ans[(2 * i)] = seq[i];
      ans[(2 * i + 1)] = val;
    }
    if (!can) {
      cout << -1;
    } else {
      for (int i = 0; i < 2 * n; i++) {
        cout << ans[i] + 1 << " ";
      }
    }
    cout << "\n";
  }
  return 0;
}