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 mod = 998244353;
const int nag = 2e5 + 5;
template <typename T>
void add(T& a, T b) {
  a += b;
  while (a >= mod) a -= mod;
  while (a < 0) a += mod;
}
template <typename T>
void mul(T& a, T b) {
  a = a * b % mod;
}
template <typename T>
void up_self(T& a, T b) {
  a = max(a, b);
}
template <typename T>
void min_self(T& a, T b) {
  a = min(a, b);
}
string find_largest_pref_palindrome(const string& a) {
  string second = a;
  reverse(second.begin(), second.end());
  second = a + "%" + second;
  int c = 0;
  vector<int> pref(2 * ((int)a.size()) + 2);
  for (int i = 1; i < second.size(); i++) {
    while (c != 0 && second[i] != second[c]) c = pref[c - 1];
    if (second[i] == second[c]) c++;
    pref[i] = c;
  }
  return second.substr(0, c);
}
long long int binexpomodulo(long long int x, long long int y) {
  long long int res = 1;
  x %= mod;
  if (!x) return 0;
  while (y) {
    if (y & 1) {
      mul(res, x);
    }
    mul(x, x);
    y >>= 1;
  }
  return res;
}
long long int nCrInOr(long long int n, long long int r) {
  long long int res = 1;
  if (r > n - r) r = n - r;
  long long int rin = 1;
  for (long long int i = 1; i <= r; i++) rin = (rin * i) % mod;
  rin = binexpomodulo(rin, mod - 2);
  for (long long int i = 1; i <= r; i++) res = (res * (n - i + 1)) % mod;
  res = (res * rin) % mod;
  return res;
}
int msb(long long int n) {
  int ans;
  for (int i = 0; i < 64; i++)
    if (n & (1LL << i)) ans = i + 1;
  return ans;
}
long long int get_(vector<long long int>& a, int pos) {
  long long int res = 0;
  for (; pos >= 0; pos = (pos & (pos + 1)) - 1) {
    res += a[pos];
  }
  return res;
}
void upd(vector<long long int>& a, int pos, int val) {
  for (; pos < a.size(); pos = (pos | (pos + 1))) a[pos] += val;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<int> b(n), a(2 * n);
    for (int i = 0; i < n; i++) cin >> b[i];
    ;
    set<int> in_b;
    for (int i : b) {
      in_b.insert(i);
    }
    vector<bool> available(2 * n + 1, true);
    bool cant = false;
    for (int i = 0; i < n; i++) {
      int c = 0;
      for (int j = b[i]; j <= 2 * n; j++) {
        if ((c == 1 && available[j] && (in_b.find(j) == in_b.end())) || (!c)) {
          available[j] = false;
          a[2 * i + c] = j;
          c++;
        }
        if (c == 2) break;
      }
      if (c < 2) {
        cant = true;
        break;
      }
    }
    if (cant)
      cout << -1 << "\n";
    else {
      for (int i = 0; i < a.size(); i++) cout << a[i] << " ";
      cout << "\n";
      ;
    }
  }
  return 0;
}