Restore the Permutation by Merger
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 $$$p[1 \dots n]$$$. 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.
The first line of the input contains one integer $$$t$$$ ($$$1 \le t \le 400$$$) — the number of test cases. Then $$$t$$$ test cases follow.The first line of the test case contains one integer $$$n$$$ ($$$1 \le n \le 50$$$) — the length of permutation. The second line of the test case contains $$$2n$$$ integers $$$a_1, a_2, \dots, a_{2n}$$$ ($$$1 \le a_i \le n$$$), 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 $$$p$$$.
For each test case, print the answer: $$$n$$$ integers $$$p_1, p_2, \dots, p_n$$$ ($$$1 \le p_i \le n$$$), representing the initial permutation. It is guaranteed that the answer exists and is unique.
t = int(input())
while t != 0:
    pl = int(input())
    p = [int(i) for i in input().split()]
    x = set()
    output = []
    for i in p:
        if i not in x:
            x.add(i)
            output.append(i)
    print(' '.join([str(i) for i in output]))
    t -= 1
for s in [*open(0)][2::2]:
    print(*dict.fromkeys(s.split()))
def permutation(n, a):
    seen = set()
    result = []
    for num in a:
        if num not in seen:
            result.append(num)
            seen.add(num)
    return result
t = int(input())
for _ in range(t):
    n = int(input())
    a = list(map(int, input().split()))
    ans = permutation(n, a)
    print(*ans)