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;
double getTime() { return clock() / (double)CLOCKS_PER_SEC; }
void read(){};
template <typename T, typename... Args>
void read(T& a, Args&... args) {
  cin >> a;
  read(args...);
}
void print(){};
template <typename T, typename... Args>
void print(T a, Args... args) {
  cout << a << " \n"[sizeof...(args) == 0];
  print(args...);
}
void solve() {
  int n;
  cin >> n;
  vector<int> a(n);
  vector<int> cnt(201, 0);
  for (int i = 0; i < n; i++) {
    read(a[i]);
    cnt[a[i]] = 1;
  }
  vector<int> ans;
  for (int i = 0; i < n; i++) {
    ans.push_back(a[i]);
    int k = 1;
    while (cnt[a[i] + k]) k++;
    ans.push_back(a[i] + k);
    cnt[a[i] + k] = 1;
    if (a[i] + k > n + n) {
      print(-1);
      return;
    }
  }
  for (auto x : ans) cout << x << ' ';
  puts("");
}
int main() {
  int t;
  read(t);
  while (t--) solve();
}