A permutation of length n is a sequence of integers from 1 to n of length n containing each number exactly once. For example, [1], [4, 3, 5, 1, 2], [3, 2, 1] are permutations, and [1, 1], [0, 1], [2, 2, 1, 4] are not.

There was a permutation p[1 ... n]. It was merged with itself. In other words, let's take two instances of p and insert elements of the second p into the first maintaining relative order of elements. The result is a sequence of the length 2n.

For example, if p=[3, 1, 2] some possible results are: [3, 1, 2, 3, 1, 2], [3, 3, 1, 1, 2, 2], [3, 1, 3, 1, 2, 2]. The following sequences are not possible results of a merging: [1, 3, 2, 1, 2, 3], [3, 1, 2, 3, 2, 1], [3, 3, 1, 2, 2, 1].

For example, if p=[2, 1] the possible results are: [2, 2, 1, 1], [2, 1, 2, 1]. The following sequences are not possible results of a merging: [1, 1, 2, 2], [2, 1, 1, 2], [1, 2, 2, 1].

Your task is to restore the permutation p by the given resulting sequence a. It is guaranteed that the answer exists and is unique.

You have to answer t independent test cases.

Input

The first line of the input contains one integer t (1 ≤ t ≤ 400) — the number of test cases. Then t test cases follow.

The first line of the test case contains one integer n (1 ≤ n ≤ 50) — the length of permutation. The second line of the test case contains 2n integers a_1, a_2, ..., a_{2n} (1 ≤ a_i ≤ n), where a_i is the i-th element of a. It is guaranteed that the array a represents the result of merging of some permutation p with the same permutation p.

Output

For each test case, print the answer: n integers p_1, p_2, ..., p_n (1 ≤ p_i ≤ n), representing the initial permutation. It is guaranteed that the answer exists and is unique.

Example

Input


5
2
1 1 2 2
4
1 3 1 4 3 4 2 2
5
1 2 1 2 3 4 3 5 4 5
3
1 2 3 1 2 3
4
2 3 2 4 1 3 4 1


Output


1 2 
1 3 4 2 
1 2 3 4 5 
1 2 3 
2 3 4 1 
"Codeforces Round #656 (Div. 3)"
"A. Three Pairwise Maximums"
# y=int(input())
# for i in range(y):
#     al=input().split()
#     m=list(map(int,al))
#     m.sort()
#     s=set(m)
#     if len(s)==3:
#         print("NO")
#     elif len(s)==1:
#         print("YES")
#         print(m[0],m[1],m[2])
#     else:
#         z=x=0
#         for j in s:
#             e=m.count(j)
#             if e==1:
#                 z=j
#             elif e==2:
#                 x=j
#         if z>x:
#             print("NO")
#         else:
#             print("YES")
#             print(m[0],m[0],m[1])            
"B. Restore the Permutation by Merger"
y=int(input())
for i in range(y):
    n=int(input())
    al=input().split()
    m=list(map(int,al))
    a=[]
    d={}
    for j in m:
        d.setdefault(j,0)
        d[j]+=1
        if d[j]==1:
            a.append(j)
    for j in a:
        print(j,end=" ",sep="")
    print()                    
