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 
from  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, .
math import inf, log2

class SegmentTree:
    def __init__(self, array, func=max):
        self.n = len(array)
        self.size = 2**(int(log2(self.n-1))+1) if self.n != 1 else 1
        self.func = func
        self.default = 0 if self.func != min else inf
        self.data = [self.default] * (2 * self.size)
        self.process(array)

    def process(self, array):
        self.data[self.size : self.size+self.n] = array
        for i in range(self.size-1, -1, -1):
            self.data[i] = self.func(self.data[2*i], self.data[2*i+1])

    def query(self, alpha, omega):
        """Returns the result of function over the range (inclusive)!"""
        if alpha == omega:
            return self.data[alpha + self.size]
        res = self.default
        alpha += self.size
        omega += self.size + 1
        while alpha < omega:
            if alpha & 1:
                res = self.func(res, self.data[alpha])
                alpha += 1
            if omega & 1:
                omega -= 1
                res = self.func(res, self.data[omega])
            alpha >>= 1
            omega >>= 1
        return res

    def update(self, index, value):
        """Updates the element at index to given value!"""
        index += self.size
        self.data[index] = value
        index >>= 1
        while index:
            self.data[index] = self.func(self.data[2*index], self.data[2*index+1])
            index >>= 1

# ------------------- fast io --------------------
import os
import sys
from io import BytesIO, IOBase

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()
           ...