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:
t=int(input())
while(t):
        t-=1
        n=int(input())
        a=list(map(int,input().split()))
        b=[0]*(2*n)
        f=set()
        flag=0
        a1=a[:]
        for i in range(n):
                if(a[i] not in f):
                        f.add(a[i])
                        b[(2*(i+1))-2]=a[i]
                else:
                        flag=1
        a.sort()
        if(flag==1):
                print(-1)
                continue
        c=[]
        for i in range(2*n):
                if(i+1 not in f):
                        c.append(i+1)
        c.sort()
        for i in range(n):
                for j in range(len(c)):
                        if(c[j]>a1[i]):
                                b[2*(i+1)-1]=c[j]
                                c.remove(c[j])
                                break
        if(b.count(0)==0):
                print(*b)
        else:
                print(-1)