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;
long long int modPow(long long int x, long long int n, long long int mod) {
  long long int r = 1;
  x = x % mod;
  while (n) {
    if (n & 1) r = (r * x) % mod;
    n = n >> 1;
    x = (x * x) % mod;
  }
  return r;
}
int main() {
  long long int t, i, j, k, l, n, m, a, b, c, x, y, z;
  cin >> t;
  while (t--) {
    cin >> n;
    vector<long long int> V;
    map<long long int, long long int> M;
    bool ans = true;
    for (i = 0; i < n; ++i) {
      cin >> l;
      M[l]++;
      V.push_back(l);
      if (M[l] > 1) ans = false;
    }
    vector<long long int> P(2 * n, 0);
    for (i = 0; i < n; ++i) P[i * 2] = V[i];
    set<pair<long long int, long long int> > S;
    for (i = 0; i < n; ++i) {
      bool milla = false;
      for (j = V[i]; j <= 2 * n; ++j) {
        if (M[j] == 0) {
          M[j]++;
          P[2 * i + 1] = j;
          milla = true;
          break;
        }
      }
      if (milla == false) {
        ans = false;
        break;
      }
    }
    if (ans) {
      for (auto it : P) cout << it << " ";
      cout << "\n";
    } else {
      cout << -1 << "\n";
    }
  }
  return 0;
}