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>
#pragma optimization_level 3
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,O3")
using namespace std;
const long long MOD = 1e+9 + 7;
const long long INF = LLONG_MAX;
const int INFi = INT_MAX;
const long long N = 3e+5 + 8;
long long a[N], b[N];
long long check[N];
long long tc = 1;
void solve(long long tc) {
  long long n;
  cin >> n;
  for (long long i = (1); i <= n; i++) cin >> a[i];
  ;
  for (long long i = (1); i <= 2 * n; i++) {
    check[i] = 0;
  }
  for (long long i = (1); i <= n; i++) {
    b[2 * i - 1] = a[i];
    check[a[i]] = 1;
  }
  for (long long i = (1); i <= n; i++) {
    long long k = b[2 * i - 1] + 1;
    while (k <= 200 && check[k] == 1) {
      k++;
    }
    b[2 * i] = k;
    check[k] = 1;
  }
  int flag = 1;
  for (long long i = (1); i <= 2 * n; i++) {
    if (check[i] == 0) {
      flag = 0;
      break;
    }
  }
  if (flag) {
    for (long long i = (1); i <= 2 * n; i++) cout << b[i] << " ";
    cout << endl;
    ;
  } else {
    cout << -1 << " ";
    cout << endl;
  };
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << fixed;
  cout << setprecision(10);
  ;
  cin >> tc;
  for (long long i = (1); i <= tc; i++) {
    solve(i);
  }
  return 0;
}