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 N = 2e2 + 10;
const long long mod = 1e9 + 7;
const long long inf = 1e18 + 10;
long long a[N], b[N], c[N], use[N];
set<int> st[N];
void solve() {
  long long n;
  cin >> n;
  fill(c, c + N, 0);
  fill(use, use + N, 0);
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    use[a[i]] = 1;
  }
  for (int i = 1; i <= n; i++)
    for (int j = a[i] + 1; j <= 2 * n; j++)
      if (!use[j]) st[i].insert(j);
  for (int i = 1; i <= n; i++) {
    if (st[i].empty()) {
      cout << -1 << '\n';
      return;
    }
    bool find = 0;
    for (int j = a[i] + 1; j <= 2 * n; j++) {
      if (st[i].find(j) == st[i].end() || use[j]) continue;
      bool nec = 0;
      for (int k = 1; k <= n; k++)
        if (k != i && c[k] == 0 && st[k].size() == 1 &&
            st[k].find(j) != st[k].end())
          nec = 1;
      if (nec) continue;
      b[i] = j;
      c[i] = 1;
      for (int k = 1; k <= n; k++)
        if (st[k].find(j) != st[k].end()) st[k].erase(j);
      find = 1;
      break;
    }
    if (find == 0) {
      cout << -1 << '\n';
      return;
    }
  }
  for (int i = 1; i <= n; i++) cout << a[i] << ' ' << b[i] << ' ';
  cout << '\n';
  for (int i = 0; i < N; i++)
    while (st[i].size()) st[i].erase(st[i].begin());
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long q;
  cin >> q;
  while (q--) solve();
  return 0;
}