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.util.*;

public class EXPCH {
	static long modInverse(long a, long m) 
	{ 
		long m0 = m; 
		long y = 0, x = 1; 

		if (m == 1) 
			return 0; 

		while (a > 1) 
		{
			long q = a / m; 

			long t = m; 

			
			m = a % m; 
			a = t; 
			t = y; 

			
			y = x - q * y; 
			x = t; 
		} 

		
		if (x < 0) 
			x += m0; 

		return x; 
	} 

	public static class pair implements Comparable<pair>{
		int idx,val;
		public pair(int i,int j) {
			idx=i;
			val=j;
		}
		@Override
		public int compareTo(pair o) {
			// TODO Auto-generated method stub
			return this.val-o.val;
		}
		
	}
	public static void main(String[] args) {
		// TODO Auto-generated method stub

		Scanner sc=new Scanner(System.in);
		int t=sc.nextInt();
	StringBuilder st=new StringBuilder();
	
		while(t-->0) {
			TreeSet<Integer>hs=new TreeSet<>();
			int n=sc.nextInt();
			int a[]=new int[2*n+1];
			pair b[]=new pair[n+1];
			b[0]=new pair(0,0);
			for(int i=1;i<=2*n;i++) {
				hs.add(i);
			}
			for(int i=1;i<=n;i++) {
				b[i]=new pair(i,sc.nextInt());
				hs.remove(b[i].val);
			}
			//Arrays.parallelSort(b);
			
			boolean f=true;
			for(int i=1;i<=n;i++) {
				int val=b[i].val;
				int idx=b[i].idx;
			
				if(hs.ceiling(val+1)==null) {
				f=false;break;
				}
				int no=hs.ceiling(val+1);
				a[idx*2]=no;
				a[idx*2-1]=val;
				hs.remove(no);
			}
			if(!f) {
				st.append(-1+"\n");
				continue;
			}
		for(int i=1;i<=2*n;i++) {
			st.append(a[i]+" ");
		}
		st.append("\n");
		}
		System.out.println(st);
	}
	


}
