human
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 
The above is tricky. Write me a correct solution in PYTHON3.
 What is a Python code to print the answer: n integers p_1, p_2,?
assistant
#Here is Python code to print the answer: n integers p_1, p_2, .

import os, sys
from io import BytesIO, IOBase
from collections import Counter
# Fast IO Region
BUFSIZE = 8192

class FastIO (IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno ()
        self.buffer = BytesIO ()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read (self._fd, max (os.fstat (self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell ()
            self.buffer.seek (0, 2), self.buffer.write (b), self.buffer.seek (ptr)
        self.newlines = 0
        return self.buffer.read ()

    def readline(self):
        while self.newlines == 0:
            b = os.read (self._fd, max (os.fstat (self._fd).st_size, BUFSIZE))
            self.newlines = b.count (b"\n") + (not b)
            ptr = self.buffer.tell ()
            self.buffer.seek (0, 2), self.buffer.write (b), self.buffer.seek (ptr)
        self.newlines -= 1
        return self.buffer.readline ()

    def flush(self):
        if self.writable:
            os.write (self._fd, self.buffer.getvalue ())
            self.buffer.truncate (0), self.buffer.seek (0)

class IOWrapper (IOBase):
    def __init__(self, file):
        self.buffer = FastIO (file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write (s.encode ("ascii"))
        self.read = lambda: self.buffer.read ().decode ("ascii")
        self.readline = lambda: self.buffer.readline ().decode ("ascii")
sys.stdin, sys.stdout = IOWrapper (sys.stdin), IOWrapper (sys.stdout)
input = lambda: sys.stdin.readline ().rstrip ("\r\n")

for _ in range(int(input())):
    x=int(input())
    y=list(map(int,input().split()))
    z=[]
    for t in y:
        if t not in z:
            z.append(t)
    print(*z)