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 <typename T>
using second = set<T>;
template <typename T>
using v = vector<T>;
const int Max = 2e5 + 100;
const int Mx = 5e1 + 10;
const int Len = 1e6;
const int MOD = 1e9 + 7;
const long long INF = 1e18;
int A[Max];
bool usedNum[Max];
int main() {
  int q;
  cin >> q;
  while (q--) {
    int n;
    cin >> n;
    bool able = true;
    fill(usedNum, usedNum + 2 * n + 100, false);
    fill(A, A + 2 * n + 100, -1);
    for (int i = 0; i < n; i++) {
      cin >> A[2 * i];
      if (!usedNum[A[2 * i]]) {
        usedNum[A[2 * i]] = true;
      } else {
        able = false;
      }
    }
    for (int i = 1; i < 2 * n; i += 2) {
      int trg = -1;
      for (int j = A[i - 1] + 1; j <= 2 * n; j++) {
        if (!usedNum[j]) {
          trg = j;
          usedNum[trg] = true;
          break;
        }
      }
      A[i] = trg;
    }
    for (int i = 0; i < 2 * n; i++) {
      if (A[i] == -1) able = false;
    }
    if (able) {
      cout << "";
      for (int i = 0; i < 2 * n; i++) {
        cout << A[i] << ' ';
      }
      cout << '\n';
    } else {
      cout << -1 << '\n';
    }
  }
}