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;
inline long long gcd(long long a, long long b) {
  if (a % b == 0)
    return b;
  else {
    return gcd(b, a % b);
  }
}
inline long long lcm(long long a, long long b) { return a * b / gcd(a, b); }
inline long long gaus(long long a, long long b) {
  return (a + b) * (b - a + 1) / 2;
}
template <class T>
ostream& operator<<(ostream& os, vector<T> v) {
  os << "[";
  int cnt = 0;
  for (auto vv : v) {
    os << vv;
    if (++cnt < v.size()) os << ",";
  }
  return os << "]";
}
template <class L, class R>
ostream& operator<<(ostream& os, map<L, R> v) {
  os << "[";
  int cnt = 0;
  for (auto vv : v) {
    os << vv;
    if (++cnt < v.size()) os << ",";
  }
  return os << "]";
}
template <class L, class R>
ostream& operator<<(ostream& os, pair<L, R> p) {
  return os << "(" << p.first << "," << p.second << ")";
}
int v[201];
void solve() {
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    memset(v, 0, sizeof(v));
    vector<int> a(n + 1);
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
      v[a[i]] = 1;
    }
    bool tof = true;
    vector<int> ans;
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= 2 * n; j++) {
        if (v[j] == 0) {
          if (a[i] < j && min(a[i], j) == a[i]) {
            ans.push_back(a[i]);
            ans.push_back(j);
            v[j] = 1;
            break;
          }
        }
      }
    }
    for (int i = 1; i <= 2 * n; i++) {
      if (v[i] == 0) tof = false;
    }
    if (tof) {
      for (int i = 0; i < ans.size(); i++) cout << ans[i] << " ";
      cout << '\n';
    } else
      cout << -1 << '\n';
  }
}
int main() {
  cin.tie(0), ios_base::sync_with_stdio(false);
  solve();
  return 0;
}