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;
int Search(vector<long long int> v1, long long int x1) {
  long long int j;
  for (j = 0; j < v1.size(); j++) {
    if (x1 == v1[j]) {
      return j;
    } else {
      continue;
    }
  }
  return -1;
}
int Search2(long long int v[], long long int x, long long int n) {
  long long int i;
  for (i = 0; i < n; i++) {
    if (x == v[i]) {
      return i;
    } else {
      continue;
    }
  }
  return -1;
}
int main() {
  long long int e, r, g, pts, y, t, d, q, x, n, h, m, w, l, k,
      cnt = 500, countA = 0, sum, sumt, countB = 0, countC = 0, a1, a2,
      countD = 0;
  long long int min = 1000;
  vector<long long int> v3;
  vector<long long int> v2;
  set<long long int> s;
  cin >> a2;
  while (a2--) {
    cin >> n;
    for (h = 0; h < n; h++) {
      cin >> sum;
      v2.push_back(sum);
    }
    e = *max_element(v2.begin(), v2.end());
    r = *min_element(v2.begin(), v2.end());
    if (e >= 2 * n || r > 1) {
      cout << -1;
    } else {
      v3.push_back(0);
      for (k = 0; k < n; k++) {
        t = v2[k];
        v3.push_back(t);
        for (r = 1; r <= (2 * n) - t; r++) {
          if (Search(v3, t + r) == -1 && Search(v2, t + r) == -1) {
            break;
          } else
            continue;
        }
        v3.push_back(t + r);
      }
      g = *max_element(v3.begin(), v3.end());
      if (g == 2 * n) {
        for (q = 1; q <= 2 * n; q++) {
          cout << v3[q] << " ";
        }
      } else {
        cout << -1;
      }
    }
    cout << "\n";
    v3.clear();
    v2.clear();
  }
  return 0;
}