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 
import sys,bisect,string,math,time,functools,random,fractions
from heapq import heappush,heappop,heapify
from collections import deque,defaultdict,Counter
from itertools import permutations,combinations,groupby
rep=range;R=range
def Golf():n,*t=map(int,open(0).read().split())
def I():return int(input())
def S_():return input()
def IS():return input().split()
def LS():return [i for i in input().split()]
def MI():return map(int,input().split())
def LI():return [int(i) for i in input().split()]
def LI_():return [int(i)-1 for i in input().split()]
def NI(n):return [int(input()) for i in range(n)]
def NI_(n):return [int(input())-1 for i in range(n)]
def StoLI():return [ord(i)-97 for i in input()]
def ItoS(n):return chr(n+97)
def LtoS(ls):return ''.join([chr(i+97) for i in ls])
def RA():return map(int,open(0).read().split())
def GI(V,E,ls=None,Directed=False,index=1):
    org_inp=[];g=[[] for i in range(V)]
    FromStdin=True if ls==None else False
    for i in range(E):
        if FromStdin:
            inp=LI()
            org_inp.append(inp)
        else:
            inp=ls[i]
        if len(inp)==2:
            a,b=inp;c=1
        else:
            a,b,c=inp
        if index==1:a-=1;b-=1
        aa=(a,c);bb=(b,c);g[a].append(bb)
        if not Directed:g[b].append(aa)
    return g,org_inp
def GGI(h,w,search=None,replacement_of_found='.',mp_def={'#':1,'.':0},boundary=1):
    #h,w,g,sg=GGI(h,w,search=['S','G'],replacement_of_found='.',mp_def={'#':1,'.':0},boundary=1) # sample usage
    mp=[boundary]*(w+2);found={}
    for i in R(h):
        s=input()
        for char in search:
            if char in s:
                found[char]=((i+1)*(w+2)+s.index(char)+1)
                mp_def[char]=mp_def[replacement_of_found]
        mp+=[boundary]+[mp_def[j] for j in s]+[boundary]
    mp+=[boundary]*(w+2)
    return h+2,w+2,mp,found
def TI(n):return GI(n,n-1)
def accum(ls):
    rt=[0]
    for i in ls:rt+=[rt[-1]+i]
    return rt
def bit_combination(n,base=2):
    rt=[]
    for tb in R(base**n):s=[tb//(base**bt)%base for bt in R(n)];rt+=[s]
    return rt
def gcd(x,y):
    if y==0:return x
    if x%y==0:return y
    while x%y!=0:x,y=y,x%y
    return y
def YN(x):print(['NO','YES'][x])
def Yn(x):print(['No','Yes'][x])
def show(*inp,end='\n'):
    if show_flg:print(*inp,end=end)

mo=10**9+7
inf=float('inf')
FourNb=[(-1,0),(1,0),(0,1),(0,-1)];EightNb=[(-1,0),(1,0),(0,1),(0,-1),(1,1),(-1,-1),(1,-1),(-1,1)];compas=dict(zip('WENS',FourNb));cursol=dict(zip('LRUD',FourNb))
l_alp=string.ascii_lowercase
#sys.setrecursionlimit(10**7)
read=sys.stdin.buffer.read;readline=sys.stdin.buffer.readline;input=lambda:sys.stdin.readline().rstrip()

########################################################################################################################################################################
########################################################################################################################################################################
# Verified by
# https://yukicoder.me/problems/no/979
# https://atcoder.jp/contests/abc152/tasks/abc152_e

## return prime factors of N as dictionary {prime p:power of p}
## within 2 sec for N = 2*10**20+7
def primeFactor(N):
    i,n=2,N
    ret={}
    d,sq=2,99
    while i<=sq:
        k=0
        while n%i==0:
            n,k,ret[i]=n//i,k+1,k+1
        if k>0 or i==97:
            sq=int(n**(1/2)+0.5)
        if i<4:
            i=i*2-1
        else:
            i,d=i+d,d^6
    if n>1:
        ret[n]=1
    return ret

## return divisors of n as list
def divisor(n):
    div=[1]
    for i,j in primeFactor(n).items():
        div=[(i**k)*d for d in div for k in range(j+1)]
    return div

## return the list of prime numbers in [2,N], using eratosthenes sieve
## around 800 ms for N = 10**6  by PyPy3 (7.3.0) @ AtCoder
def PrimeNumSet(N):
    M=int(N**0.5)
    seachList=[i for i in range(2,N+1)]
    primes=[]
    while seachList:
        if seachList[0]>M:
            break
        primes.append(seachList[0])
        tmp=seachList[0]
        seachList=[i for i in seachList if i%tmp!=0]
    return primes+seachList


## retrun LCM of numbers in list b
## within 2sec for no of B = 10*5  and  Bi < 10**6
def LCM(b,mo=10**9+7):
    prs=PrimeNumSet(max(b))
    M=dict(zip(prs,[0]*len(prs)))
    for i in b:
        dc=primeFactor(i)
        for j,k in dc.items():
            M[j]=max(M[j],k)
    
    r=1
    for j,k in M.items():
        if k!=0:
            r*=pow(j,k,mo)
            r%=mo
    return r

show_flg=False
show_flg=True

ans=0
for _ in range(I()):
    n=I()
    a=LI()
    ans=[]
    v=[0]*n
    for i in range(2*n):
        if v[a[i]-1]==0:
            v[a[i]-1]=1
            ans+=a[i],
    
    print(*ans)



