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 
import java.io.*;
import java.util.Arrays;
import java.util.BitSet;
import java.util.Comparator;
import java.util.Scanner;

import static java.lang.Math.*;

public class C {
    private static class Pair<K, V> {
        K k;
        V v;

        public Pair(K k, V v) {
            this.k = k;
            this.v = v;
        }
    }

    public static void main(String[] args) {
        Scanner in = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
        PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
        int t = in.nextInt();
        for (int z = 0; z < t; z++) {
            int n = in.nextInt();
            Pair<Pair<Integer, Integer>, Pair<Integer, Integer>>[] arr = new Pair[n];
            BitSet used = new BitSet();
            for (int i = 0; i < n; i++) {
                int a = in.nextInt();
                used.set(a, true);
                arr[i] = new Pair<>(new Pair<>(a, i), new Pair<>(a, a));
            }
            //Arrays.sort(arr, Comparator.comparingInt((o) -> o.k.k));
            boolean bad = false;
            for (int i = 0; i < n; i++) {
                Pair<Pair<Integer, Integer>, Pair<Integer, Integer>> p = arr[i];
                int f = used.nextClearBit(p.k.k);
                if (f > n * 2) {
                    out.println("-1");
                    bad = true;
                    break;
                }
                p.v.k = p.k.k;
                p.v.v = f;
                used.set(f);
            }
            if (bad) continue;
            //Arrays.sort(arr, Comparator.comparingInt((o) -> o.k.v));
            for (int i = 0; i < n; i++) {
                Pair<Pair<Integer, Integer>, Pair<Integer, Integer>> p = arr[i];
                int r = p.k.k;

            }

            for (int i = 0; i < n; i++) out.print(arr[i].v.k + " " + arr[i].v.v + " ");
            out.println();
        }
        out.close();
    }
}
