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;
const int inf = 1e9 + 7;
const long long INF = 1e18;
template <typename T1, typename T2>
inline bool chmin(T1& a, T2 b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
template <typename T1, typename T2>
inline bool chmax(T1& a, T2 b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <typename T>
T gcd(T a, T b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
template <typename T>
T lcm(T a, T b) {
  return a / gcd(a, b) * b;
}
template <typename T>
T pow(T a, int b) {
  return b ? pow(a * a, b / 2) * (b % 2 ? a : 1) : 1;
}
const int mod = 1000000007;
long long modpow(long long a, int b) {
  return b ? modpow(a * a % mod, b / 2) * (b % 2 ? a : 1) % mod : 1;
}
template <class T>
ostream& operator<<(ostream& os, const vector<T>& vec) {
  for (int i = 0; i < ((int)vec.size()); ++i) {
    if (i) os << " ";
    os << vec[i];
  }
  return os;
}
template <class T, class U>
ostream& operator<<(ostream& os, const pair<T, U>& p) {
  os << p.first << " " << p.second;
  return os;
}
template <class T>
inline void add(T& a, int b) {
  a += b;
  if (a >= mod) a -= mod;
}
void solve();
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int T;
  cin >> T;
  while (T--) {
    solve();
  }
}
void solve() {
  int n;
  cin >> n;
  vector<int> b(n);
  for (int i = 0; i < (n); ++i) cin >> b[i];
  vector<int> a(2 * n);
  for (int i = 0; i < (n); ++i) a[2 * i] = b[i];
  set<int> t;
  for (int i = 0; i < (n); ++i) t.insert(b[i]);
  set<int> s;
  for (int i = 0; i < (2 * n); ++i)
    if (!t.count(i + 1)) s.insert(i + 1);
  for (int i = 0; i < (n); ++i) {
    auto it = s.upper_bound(b[i]);
    if (it == s.end()) {
      cout << -1 << endl;
      return;
    }
    a[2 * i + 1] = *it;
    s.erase(it);
  }
  cout << a << endl;
}