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>
#pragma GCC optimize("O3")
#pragma GCC target("sse4")
using namespace std;
template <class T>
bool ckmin(T& a, const T& b) {
  return b < a ? a = b, 1 : 0;
}
template <class T>
bool ckmax(T& a, const T& b) {
  return a < b ? a = b, 1 : 0;
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int mod = 1e9 + 7;
const char nl = '\n';
void solve() {
  int t;
  cin >> t;
  vector<int> b, a;
  unordered_map<int, int> all;
  while (t--) {
    int n;
    cin >> n;
    b.resize(n);
    all.clear();
    bool all_unique = true;
    for (int i = 0; i < n; ++i) {
      cin >> b[i];
      if (all.find(b[i]) != all.end()) {
        all_unique = false;
      }
      all[b[i]] = i;
    }
    if (!all_unique) {
      cout << -1 << "\n";
      continue;
    }
    a.assign(2 * n, -1);
    bool good = true;
    for (int m = 1; m <= 2 * n; ++m) {
      if (all.find(m) != all.end()) {
        int pos = all[m];
        a[2 * pos] = m;
      } else {
        bool found = false;
        int ind = 0;
        for (int i = 0; i < n; ++i) {
          if (a[2 * i] != -1 && a[2 * i + 1] == -1) {
            found = true;
            ind = i;
            break;
          }
        }
        if (found) {
          a[2 * ind + 1] = m;
        } else {
          good = false;
          break;
        }
      }
    }
    if (!good) {
      cout << -1;
    } else {
      for (int i = 0; i < 2 * n; ++i) {
        cout << a[i] << " ";
      }
    }
    cout << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  solve();
  return 0;
}