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:
import bisect
t=int(input())

def find_gt(a, x):
    'Find leftmost value greater than x'
    i = bisect.bisect_right(a, x)
    if i != len(a):
        return a[i]
    raise ValueError
for _ in range(t):
    b=int(input())
    B=list(map(int,input().split()))
    d={}
    v=[0]
    x=[]
    y=[]
    for i in range(1,2*b+1):
        v.append(0)
    for i in B:
        v[i]=1
        
    for i in range(1,2*b+1):
        if(v[i]==0):
            y.append(i)
        if(v[i]):
            x.append(i)

    i=0
    f=1
    while(i<b):
        if(x[i]<y[i]):
            d[x[i]]=y[i]
        else:
            print(-1)
            f=0
            break
        i+=1
    if(f):
        for i in B:
            s=find_gt(y,i)
            y.remove(s)
            print(i,s,end=" ")
        print()