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 T1, typename T2>
istream &operator>>(istream &is, vector<pair<T1, T2>> &v) {
  for (pair<T1, T2> &t : v) is >> t.first >> t.second;
  return is;
}
template <typename T>
istream &operator>>(istream &is, vector<T> &v) {
  for (T &t : v) is >> t;
  return is;
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  for (const T &t : v) {
    os << t << " ";
  }
  os << '\n';
  return os;
}
double pi = acos(-1.0);
long long md = 1000000007;
long long abst(long long a) { return a < 0 ? -a : a; }
struct HASH {
  size_t operator()(const pair<long long, long long> &x) const {
    return (size_t)x.first * 37U + (size_t)x.second;
  }
};
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
long long Pow(long long n, long long x) {
  long long ans = 1;
  while (x > 0) {
    if (x & 1) ans = (ans * n) % md;
    n = (n * n) % md;
    x = x >> 1;
  }
  return ans;
}
vector<long long> fact, inv;
void inverse(long long n) {
  inv.resize(n + 1);
  inv[0] = 1;
  for (long long i = 1; i <= n; i++) inv[i] = Pow(fact[i], md - 2);
}
void factorial(long long n) {
  fact.resize(n + 1);
  fact[0] = 1;
  for (long long i = 1; i <= n; i++) fact[i] = (fact[i - 1] * i) % md;
}
long long max(long long a, long long b) { return a > b ? a : b; }
long long min(long long a, long long b) { return a < b ? a : b; }
void solve() {
  long long n;
  cin >> n;
  vector<long long> v(n);
  cin >> v;
  vector<long long> ans(2 * n);
  vector<long long> vis(2 * n + 1, -1);
  for (long long i = 0; i < n; i++) {
    ans[2 * i] = v[i];
    vis[v[i]] = 1;
  }
  long long ptr;
  for (long long i = 1; i < 2 * n; i += 2) {
    ptr = -1;
    for (long long j = 1; j <= 2 * n; j++)
      if (vis[j] == -1 && j > ans[i - 1]) {
        ptr = j;
        break;
      }
    if (ptr == -1) {
      cout << -1 << "\n";
      return;
    }
    ans[i] = ptr;
    vis[ptr] = 1;
  }
  cout << ans;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}