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:


"""
NTC here
""" 
import sys
inp= sys.stdin.readline
input = lambda : inp().strip()
# flush= sys.stdout.flush
# import threading
# sys.setrecursionlimit(10**6)
# threading.stack_size(2**26)

def iin(): return int(input())
def lin(): return list(map(int, input().split()))



def main():
    T = iin()
    while T:
        T-=1
        n = iin()
        a = lin()
        l = 2*n+1
        done = [0]*(l)
        ans = []
        for i in a:
            done[i]=1
        for i in a:
            ans.append(i)
            for j in range(i+1, l):
                if done[j]==0:
                    done[j]=1
                    ans.append(j)
                    break
            else:
                print(-1)
                break
        else:
            print(*ans)



main()

#threading.Thread(target=main).start()