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 inf = 0x3f3f3f3f;
const int maxn = 100 + 10;
int n;
int b[maxn];
int ans[2 * maxn];
void solve() {
  cin >> n;
  map<int, int> m;
  m.clear();
  for (int i = 1; i <= n; i++) {
    cin >> b[i];
    m[b[i]] = 1;
    ans[2 * i - 1] = b[i];
  }
  for (int i = 1; i <= n; i++) {
    for (int j = b[i] + 1;; j++) {
      if (j > 2 * n) {
        puts("-1");
        return;
      }
      if (m[j] == 0) {
        m[j] = 1;
        ans[2 * i] = j;
        break;
      }
    }
  }
  cout << ans[1];
  for (int i = 2; i <= 2 * n; i++) cout << ' ' << ans[i];
  cout << endl;
}
int main() {
  long long T;
  scanf("%lld\n", &T);
  while (T--) solve();
  return 0;
}