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 maxn = 1e5 + 9;
const int MOD = 1000000007;
int mod = 999998639;
inline long long qpow(long long b, long long e, long long m = MOD) {
  long long a = 1;
  for (; e; e >>= 1, b = b * b % m)
    if (e & 1) a = a * b % m;
  return a;
}
set<int> s;
int b[109];
vector<int> ve;
int main() {
  int _;
  while (scanf("%d", &_) != EOF) {
    while (_--) {
      ve.clear();
      s.clear();
      int n;
      scanf("%d", &n);
      for (int i = 1; i <= 2 * n; i++) {
        s.insert(i);
      }
      for (int i = 1; i <= n; i++) {
        scanf("%d", &b[i]);
        s.erase(b[i]);
      }
      bool flag = false;
      for (int i = 1; i <= n; i++) {
        set<int>::iterator it = s.upper_bound(b[i]);
        if (it == s.end()) {
          flag = true;
          break;
        } else {
          ve.push_back(b[i]);
          ve.push_back(*it);
          s.erase(it);
        }
      }
      if (flag) {
        printf("-1\n");
      } else {
        for (int i = 0; i < ve.size(); i++) {
          printf("%d ", ve[i]);
        }
        printf("\n");
      }
    }
  }
}