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 <class T>
ostream &operator<<(ostream &os, const vector<T> &V) {
  os << "[ ";
  for (auto v : V) os << v << " ";
  return os << "]";
}
template <class T>
ostream &operator<<(ostream &os, const set<T> &S) {
  os << "{ ";
  for (auto s : S) os << s << " ";
  return os << "}";
}
template <class T>
ostream &operator<<(ostream &os, const unordered_set<T> &S) {
  os << "{ ";
  for (auto s : S) os << s << " ";
  return os << "}";
}
template <class L, class R>
ostream &operator<<(ostream &os, const pair<L, R> &P) {
  return os << "(" << P.first << "," << P.second << ") ";
}
template <class L, class R>
ostream &operator<<(ostream &os, const map<L, R> &M) {
  os << "{ ";
  for (auto m : M) os << "(" << m.first << ":" << m.second << ") ";
  return os << "}";
}
template <class T>
ostream &operator<<(ostream &os, stack<T> S) {
  while (!S.empty()) {
    os << S.top() << "\n";
    S.pop();
  }
  return os;
}
template <class T>
ostream &operator<<(ostream &os, queue<T> Q) {
  while (!Q.empty()) {
    os << Q.front() << " ";
    Q.pop();
  }
  return os;
}
const long long mod = 1e9 + 7;
inline long long modulo(long long x) {
  x = x % mod;
  return x < 0 ? x + mod : x;
}
inline long long add(long long x, long long y) {
  long long z = modulo(x) + modulo(y);
  return modulo(z);
}
inline long long sub(long long x, long long y) {
  long long z = modulo(x) - modulo(y);
  return modulo(z);
}
inline long long mul(long long x, long long y) {
  long long z = modulo(x) * 1ll * modulo(y);
  return modulo(z);
}
inline long long power(long long x, long long y) {
  x = modulo(x);
  y = modulo(y);
  if (y == 0) return 1;
  long long temp = power(x, y / 2);
  temp = mul(temp, temp);
  if (y % 2 == 1) temp = mul(temp, x);
  return temp;
}
inline long long inv(long long x) { return power(x, mod - 2); }
void solve() {
  long long n;
  cin >> n;
  vector<long long> cache(2 * n, 0), dp(2 * n + 1, 0);
  for (long long i = 0; i < n; i++) {
    cin >> cache[2 * i];
    dp[cache[2 * i]] = 1;
  }
  long long count = 0;
  for (long long i = 1; 1 < 2 * n ? i <= 2 * n : i >= 2 * n;
       1 < 2 * n ? i++ : i--) {
    if (dp[i] == 0 && count < (i + 1) / 2) {
      cout << -1 << endl;
      return;
    }
    if (dp[i] == 1) count++;
  }
  for (long long i = 0; i < n; i++) {
    long long index = cache[2 * i];
    while (dp[index]) index++;
    cache[2 * i + 1] = index;
    dp[index] = 1;
    index++;
  }
  for (auto v : cache) cout << v << " ";
  cout << endl;
}
int32_t main() {
  long long T = 1;
  cin >> T;
  for (long long t = 1; 1 < T ? t <= T : t >= T; 1 < T ? t++ : t--) {
    solve();
  }
  return 0;
}