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;
double tick() {
  static clock_t oldtick;
  clock_t newtick = clock();
  double diff = 1.0 * (newtick - oldtick) / CLOCKS_PER_SEC;
  oldtick = newtick;
  return diff;
}
long long gcd(long long a, long long b) {
  if ((a == 0) || (b == 0)) {
    return a + b;
  }
  return gcd(b, a % b);
}
long long powMod(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (res * a) % 1000000007;
    a = (a * a) % 1000000007;
    b >>= 1;
  }
  return (res % 1000000007);
}
long long pow2(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (res * a);
    a = (a * a);
    b >>= 1;
  }
  return res;
}
bool isPrime(long long a) {
  for (long long i = 3; (i * i) <= a; i += 2) {
    if ((a % i) == 0) return false;
  }
  if ((a != 2) && ((a % 2) == 0)) {
    return false;
  }
  if (a == 1) {
    return false;
  }
  return true;
}
string conLlToStr(long long a) {
  stringstream mystr;
  mystr << a;
  return mystr.str();
}
long long conStrToLl(string st) {
  long long numb = 0;
  long long len = st.size(), i, j = 0;
  for (long long i = len - 1; i >= 0; i--) {
    numb += (pow2(10, j) * (st[i] - '0'));
    j++;
  }
  return numb;
}
long long basenoToDecimal(string st, long long basee) {
  long long i = 0, j, anss = 0;
  for (long long j = (int)st.length() - 1; j >= 0; j--) {
    anss += (st[j] - '0') * pow2(basee, i);
    i++;
  }
  return anss;
}
string decimalToString(long long num, long long basee) {
  long long i = 0, j, opop;
  string anss = "";
  vector<string> stri;
  stri.push_back("0");
  stri.push_back("1");
  stri.push_back("2");
  stri.push_back("3");
  stri.push_back("4");
  stri.push_back("5");
  stri.push_back("6");
  stri.push_back("7");
  stri.push_back("8");
  stri.push_back("9");
  stri.push_back("A");
  stri.push_back("B");
  stri.push_back("C");
  stri.push_back("D");
  stri.push_back("E");
  stri.push_back("F");
  if (num == 0) {
    return "0";
  }
  while (num) {
    opop = (num % basee);
    anss += stri[opop];
    num /= basee;
  }
  reverse(anss.begin(), anss.end());
  return anss;
}
long long my_sqrt(long long x) {
  long long y = (long long)(sqrtl((long double)x) + 0.5);
  while (y * y < x) {
    y++;
  }
  while (y * y > x) {
    y--;
  }
  if (y * y == x) {
    return y;
  }
  return -1;
}
long long my_crt(long long x) {
  long long y = (long long)(powl((long double)x, 1.0 / 3.0) + 0.5);
  while (y * y * y < x) {
    y++;
  }
  while (y * y * y > x) {
    y--;
  }
  if (y * y * y == x) {
    return y;
  }
  return -1;
}
const int dx[] = {0, 1, 1, 1, 0, -1, -1, -1};
const int dy[] = {1, 1, 0, -1, -1, -1, 0, 1};
long long arr[110];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long T;
  cin >> T;
  while (T--) {
    long long N;
    cin >> N;
    for (long long i = 1; i <= N; i++) {
      cin >> arr[i];
    }
    set<long long> se;
    for (long long i = 1; i <= 2 * N; i++) {
      se.insert(i);
    }
    for (long long i = 1; i <= N; i++) {
      se.erase(arr[i]);
    }
    vector<long long> ans;
    for (long long i = 1; i <= N; i++) {
      auto it = se.lower_bound(arr[i]);
      if (it != se.end()) {
        ans.push_back(arr[i]);
        ans.push_back(*it);
        se.erase(it);
      } else {
        break;
      }
    }
    if (ans.size() == (2 * N)) {
      for (long long i = 0; i <= (int)ans.size() - 1; i++) {
        cout << ans[i] << " ";
      }
    } else {
      cout << "-1";
    }
    cout << "\n";
  }
  return 0;
}