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 
t = int(input())
for T in range(t):
    n = int(input())
    b = list(map(int,input().split()))

    ans_arr = []
    ans = 0 
    for i in range(len(b)):
        # ans_arr.append(b[i])
        # # print(ans_arr)
        # for j in range(1,2*n):
        #     if b[i] + j not in  b and b[i] + j not in ans_arr:
        #         ans_arr.append(b[i]+j)
                
                # break
        if b[i]<=2*n:
            ans_arr.append(b[i])
            for j in range(1,2*n+1):
                if b[i] + j not in  b and b[i] + j not in ans_arr:
                    ans_arr.append(b[i]+j)
                    
                    break
        else:
            ans = -1
            break


     
    # print(ans_arr)
    for i in range(1,2*n):
        if len(ans_arr)!= 0 and  i not in ans_arr:
            ans = -1
            break
    if ans == -1:
        print(-1)
    else:
        print(*ans_arr)