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;
template <typename T>
inline T gcd(T a, T b) {
  while (b != 0) swap(b, a %= b);
  return a;
}
template <typename T>
inline void seethis(vector<T> vect) {
  for (T x : vect) cout << x << " ";
  cout << "\n";
}
void err(istream_iterator<string> it) {}
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " = " << a << endl;
  err(++it, args...);
}
int lcm(int a, int b) { return a * (b / gcd(a, b)); }
bool cmp(const pair<int, int> &a) { return a.first < a.second; }
long long int modpower(long long int a, long long int b, long long int c) {
  long long int res = 1;
  while (b) {
    if (b & 1LL) res = (res * a) % c;
    a = (a * a) % c;
    b >>= 1;
  }
  return res;
}
int main() {
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(false);
  ;
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<int> b(n);
    vector<int> ans(2 * n);
    vector<bool> aval(2 * n + 1, true);
    for (int i = 0; i < n; i++) {
      cin >> b[i];
      aval[b[i]] = false;
      ans[2 * i] = b[i];
    }
    bool flag = true;
    for (int i = 1; i < 2 * n; i += 2) {
      int x = -1;
      for (int j = ans[i - 1] + 1; j <= 2 * n; j++)
        if (aval[j]) {
          x = j;
          aval[j] = false;
          break;
        }
      if (x == -1) {
        flag = false;
        break;
      }
      ans[i] = x;
    }
    if (flag) {
      seethis(ans);
    } else {
      cout << "-1\n";
    }
  }
  return 0;
}