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;
long long gcd(long long n, long long m) {
  if (n == 0) return m;
  return gcd(m % n, n);
}
bool ifprime(long long n) {
  if (n == 1) return false;
  long long i;
  for (i = 2; i * i <= n; i++) {
    if (n % i == 0) return false;
  }
  return true;
}
void disp(vector<long long> v) {
  for (auto u : v) cout << u << " ";
  cout << "\n";
}
void remleadzero(vector<int> &v) {
  vector<int> u;
  int i = 0;
  while (v[i] == 0) i++;
  while (i < v.size()) {
    u.push_back(v[i]);
    i++;
  }
  v = u;
}
long long fast_expo(long long n, long long m, long long md) {
  long long a = 1;
  while (m > 0) {
    if (m & 1) a = (a * n) % md;
    n = (n * n) % md;
    m /= 2;
  }
  return a % md;
}
long long sqroot(long long n) {
  long double N = n;
  N = sqrtl(N);
  long long sq = N - 2;
  sq = max(sq, 0LL);
  while (sq * sq < n) {
    sq++;
  }
  if ((sq * sq) == n) return sq;
  return -1;
}
int dig_sum(long long n) {
  int ret = 0;
  while (n > 0) {
    ret += (n % 10);
    n /= 10;
  }
  return ret;
}
int main() {
  ios::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  ;
  long long t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<int> v;
    for (int i = 1; i <= 2 * n; i++) v.push_back(i);
    int b[n];
    for (int i = 0; i < n; i++) {
      cin >> b[i];
      if (count((v).begin(), (v).end(), b[i])) {
        int j = 0;
        for (; j < v.size(); j++) {
          if (v[j] == b[i]) break;
        }
        v.erase(v.begin() + j);
      }
    }
    sort((v).begin(), (v).end());
    vector<int> ans;
    int done = 0;
    for (int i = 0; i < n; i++) {
      ans.push_back(b[i]);
      int j, found = 0;
      for (j = 0; j < v.size(); j++) {
        if (v[j] > b[i]) {
          found = 1;
          break;
        }
      }
      if (!found) {
        done = 1;
        break;
      }
      ans.push_back(v[j]);
      v.erase(v.begin() + j);
    }
    if (done) {
      cout << -1 << '\n';
      continue;
    }
    for (int i = 0; i < n; i++) {
      if (min(ans[2 * i], ans[2 * i + 1]) != b[i]) {
        cout << -1;
        done = 1;
        break;
      }
    }
    if (!done) {
      for (auto u : ans) cout << u << ' ';
    }
    cout << '\n';
  }
  return 0;
}