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.
assistant
#Here is Python program to print the answer: n integers p_1, p_2, .

"""
// Author : snape_here - Susanta Mukherjee
     
 """
 
from __future__ import division, print_function
 
import os,sys
from io import BytesIO, IOBase
 
if sys.version_info[0] < 3:
    from __builtin__ import xrange as range
    from future_builtins import ascii, filter, hex, map, oct, zip
 
def ii(): return int(input())
def fi(): return float(input())
def si(): return input()
def msi(): return map(str,input().split())
def mi(): return map(int,input().split())
def li(): return list(mi())
 
 
def read():
    sys.stdin = open('input.txt', 'r')  
    sys.stdout = open('output.txt', 'w') 
 
def gcd(x, y):
    while y:
        x, y = y, x % y
    return x

mod=1000000007 

def modInverse(b,m): 
    g = gcd(b, m)  
    if (g != 1):         
        return -1
    else:          
        return pow(b, m - 2, m) 

def modu(a,b,m): 

    a = a % m 
    inv = modInverse(b,m) 
    if(inv == -1): 
        return -999999999
    else: 
        return (inv*a)%m

from math import log,sqrt,factorial,cos,tan,sin,radians,ceil,floor

import bisect

from decimal import *

getcontext().prec = 25

abc="abcdefghijklmnopqrstuvwxyz"

pi=3.141592653589793238

def main():

    for _ in range(ii()):
        n=ii()
        a=li()
        l=[]
        f=[0]*51
        for i in range(2*n):
            if f[a[i]]:
                continue 
            else:
                f[a[i]]=1 
                l.append(a[i])
        print(*l)

# region fastio
 
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
       ...