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;
bool cmp(pair<long long int, long long int> a,
         pair<long long int, long long int> b) {
  return a.first < b.first;
}
bool check(int a) {
  int sum = 0;
  while (a > 0) {
    sum += (a % 10);
    a /= 10;
  }
  if (sum == 10) return true;
  return false;
}
bool prime[10005];
void seivertexe() {
  memset(prime, 1, sizeof(prime));
  prime[0] = 0;
  prime[1] = 0;
  for (long long int i = 2; i <= 10000; i++) {
    if (prime[i] == 1) {
      for (long long int j = i * i; j <= 10000; j += i) prime[j] = 0;
    }
  }
}
template <typename T>
std::string NumberToString(T Number) {
  std::ostringstream ss;
  ss << Number;
  return ss.str();
}
bool isPrime(long long int n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long int i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
long long int power(long long int x, long long int y) {
  if (y == 0)
    return 1;
  else if (y % 2 == 0)
    return (power(x, y / 2) % 1000000007 * power(x, y / 2) % 1000000007) %
           1000000007;
  else
    return (x % 1000000007 *
            (power(x, y / 2) % 1000000007 * power(x, y / 2) % 1000000007) %
            1000000007) %
           1000000007;
}
long long int factorial1(long long int n) {
  return (n == 1 || n == 0)
             ? 1
             : (n % 1000000007 * factorial1(n - 1) % 1000000007) % 1000000007;
}
long long int fact(long long int n);
long long int nCr(long long int n, long long int r) {
  return fact(n) % 1000000007 /
         ((fact(r) % 1000000007 * fact(n - r) % 1000000007)) % 1000000007;
}
long long int fact(long long int n) {
  long long int res = 1;
  for (int i = 2; i <= n; i++) res = (res * i) % 1000000007;
  return res % 1000000007;
}
void solve() {}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long int t;
  cin >> t;
  while (t--) {
    long long int n;
    cin >> n;
    long long int arr[n + 1];
    map<long long int, long long int> mp1, mp2, mp3;
    for (long long int i = 1; i <= n; i++) {
      cin >> arr[i];
      mp1[arr[i]]++;
    }
    long long int k = 0;
    long long int flag = 0;
    long long int arr1[2 * n + 2];
    for (long long int i = 1; i <= n; i++) {
      long long int x = arr[i];
      if (x > 2 * n) {
        flag = 1;
        cout << -1 << "\n";
        break;
      }
      if (mp2[x] > 0) {
        flag = 1;
        cout << -1 << "\n";
        break;
      }
      arr1[k++] = arr[i];
      long long int y = arr[i] + 1;
      while (mp1[y] != 0 || mp2[y] != 0) {
        y++;
      }
      if (y > 2 * n) {
        flag = 1;
        cout << -1 << "\n";
        break;
      } else {
        mp2[y]++;
        mp2[x]++;
        arr1[k++] = y;
      }
    }
    if (!flag) {
      for (long long int i = 0; i < 2 * n; i++) {
        cout << arr1[i] << " ";
      }
      cout << "\n";
    }
  }
  return 0;
}