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;
vector<long long> vprime;
void SieveOfEratosthenes(int n) {
  bool prime[n + 1];
  memset(prime, true, sizeof(prime));
  for (int p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      for (int i = p * p; i <= n; i += p) prime[i] = false;
    }
  }
  for (int p = 2; p <= n; p++)
    if (prime[p]) vprime.push_back(p);
}
int countFreq(string &pat, string &txt) {
  int M = pat.length();
  int N = txt.length();
  int res = 0;
  for (int i = 0; i <= N - M; i++) {
    int j;
    for (j = 0; j < M; j++)
      if (txt[i + j] != pat[j]) break;
    if (j == M) {
      res++;
      j = 0;
    }
  }
  return res;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long a, b, c, d, e, g, h, k, l, m, n, t, f,
      f1 = 0, flag2 = 0, flag = 0, flag1 = 0, co = 0, co1 = 0, co2 = 0, sum = 0,
      sum1 = 0, ma = 0, ma1 = 0, ma2 = 0, mi = 100000000000, mi1 = 100000000000,
      mi2 = 100, mi3 = 1000;
  long long co3 = 0, co4 = 0, co5 = 0, co6 = 0, co7 = 0, co8 = 0, mul = 1,
            sum2 = 0;
  long long arr[100001] = {0}, arr1[100001] = {0};
  long long a1, a2, b1, b2, c1, c2, a3, a4, b3, b4, b5, b6, m1, m2, k1, l1, m3,
      m4, d1, d2;
  double pi = 2 * acos(0.0);
  char ch;
  string str, str1, str2, str3, str4, str5;
  set<long long> s1;
  set<long long> s2;
  set<long long> s3;
  stack<long long> st;
  int x[] = {1, 0, 0, -1};
  int y[] = {0, 1, -1, 0};
  cin >> t;
  while (t--) {
    cin >> n;
    vector<long long> v(n);
    vector<long long> v1, v2, v3;
    map<long long, long long> mp, mp1;
    for (int i = 0; i < n; i++) {
      cin >> v[i];
      mp[v[i]] = 1;
      v1.push_back(v[i]);
    }
    for (int i = 1; i <= 2 * n; i++) {
      if (mp[i] == 0) {
        v2.push_back(i);
      }
    }
    sort(v1.begin(), v1.end());
    sort(v2.begin(), v2.end());
    for (int i = 0; i < v1.size(); i++) {
      mp1[v1[i]] = v2[i];
    }
    flag = 0;
    for (int i = 0; i < v.size(); i++) {
      if (mp1[v[i]] > v[i]) {
        v3.push_back(v[i]);
        v3.push_back(mp1[v[i]]);
      } else {
        flag = 1;
        break;
      }
    }
    if (flag == 1) {
      cout << -1 << endl;
      continue;
    }
    for (int i = 1; i < v3.size(); i += 2) {
      for (int j = i; j < v3.size(); j += 2) {
        if (v3[i] > v3[j]) {
          if (v3[j] > v3[i - 1]) {
            k = v3[i];
            v3[i] = v3[j];
            v3[j] = k;
          }
        }
      }
    }
    for (int i = 0; i < v3.size(); i++) cout << v3[i] << " ";
    cout << endl;
  }
  return 0;
}