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:
if __name__ == '__main__':
	for _ in range (int(input())):
		n = int(input())
		l = list(map(int, input().split()))
		a = min(l)
		b = max(l)
		if a != 1 or b == 2*n:
			print(-1)
		else:
			B = []
			d = dict()
			for i in range (n):
				d.setdefault(l[i],1)
			for i in range (1,(2*n)+1):
				d.setdefault(i,0)
				if d[i] == 0:
					B.append(i)
			B.sort()
			ans = []
			i = 0
			while i < n:
				ans.append(l[i])
				for j in range (len(B)):
					if B[j] > l[i]:
						ans.append(B[j])
						B.pop(j)
						break
				i+=1
			if len(ans) < 2*n:
				print(-1)
				continue
			print(*ans)