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 PYTHON3):

Solution:
#Ashish Sagar
import math
q=int(input())
#q=1
for _ in range(q):
    n=int(input())
    b=list(map(int,input().split()))
    ans=[0]*(2*n)
    if min(b)!=1 or max(b)==2*n:
        print(-1)
    else:
        l=[1]*(2*n+1)
        for i in range(n):
            l[b[i]]=0
        #print(l)    
        '''aa=[]
        for i in range(1,2*n+1):
            if l[i]==1:
                aa.append(i)'''
        #print(aa)        
        k=0
        for i in range(2*n):
            if i%2==0:
                ans[i]=b[k]
                k+=1
        l[0]=0        
        #print(l)        
        for i in range(0,2*n,2):
            for j in range(ans[i]+1,2*n+1):
                if l[j]==1:
                    ans[i+1]=j
                    l[j]=0
                    break
        if ans.count(0)!=0:
            print(-1)
        else:
            print(*ans)