Problem: 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 

Solution (in PYTHON3):

Solution:
t=int(input())
for _ in range(t):
    n=int(input())
    l=list(map(int,input().split()))
    out=[]
    for i in l:
        if i in out:
            continue
        else:
            out.append(i)
    for i in out:
        print(i,end=' ')
    print()
        















'''
t=int(input())
for _ in range(t):
    x,y,z=map(int,input().split())
    l=[x,y,z]
    if l.count(max(l))<2:
        print('NO')
    else:
        print('YES')
        mi=min(l)
        mx=max(l)
        idx=l.index(mi)
        if idx==0:
            print(mi,mi,mx)
        elif idx==1:
            print(mx,mi,mi)
        else:
            print(mi,mx,mi)
'''


'''
import math 
def pf(n):
    a=[1]
    while n % 2 == 0: 
        a.append(2)
        n = n / 2
    for i in range(3,int(math.sqrt(n))+1,2): 
        while n % i== 0:
            a.append(i) 
            n = n / i 
    if n > 2: 
        a.append(n)
    return sorted(a)


t=int(input())
for _ in range(t):
    n=int(input())
    if n%2==0:
        print(n//2,n//2)
    else:
        sol=pf(n)
        if sol[-1]==n:
            out=sol[-2]
        else:
            out=sol[-1]
        
        print(int(out),int(n-out))
'''
'''
t=int(input())
for _ in range(1,t+1):
    n=int(input())
    l=list(map(int,input().split()))
    s=['A','B','C','D']
    m=l[0]
    ml=[0]
    mil=[0]
    mi=l[0]
    cm=0
    cmi=0
    for i in range(n):
        if l[i]>m:
            m=l[i]
            cm+=1
            #print(cm)
            ml.append(cm)
        else:
            cm=0
            m=l[i]
            #print(cm)
        if l[i]<mi:
            mi=l[i]
            cmi+=1
            mil.append(cmi)
        else:
            cmi=0
            mi=l[i]
    ma=max(ml)+1
    mi=max(mil)+1
    ma=(ma//4)-1
    mi=(mi//4)-1
    out=0
    for i in (ma,mi):
        if i>0:
            out+=i
    print('Case #{}: {}'.format(_,out))
'''
'''
def f(k, arr):
    if k<0:
        return 0
    if k==0:
        return 1
    for j in arr:
        t=f(k-j,arr)
        if t:
            return k
    return 0
def unboundedKnapsack(k, arr):
    if k<0:
        return 0
    if k==0:
        return 1
    for i in range(k,-1,-1):
        o=f(i,arr)
        #print(i,o)
        if o:
            return i
t=int(input())
out=[]
for _ in range(t):
    n,k=map(int,input().split())
    arr=list(map(int,input().split()))
    out.append(unboundedKnapsack(k,arr))
for i in out:
    print(i)
'''
'''
n=17048
k=99
x=1
#761 99 1
#17048 14319 1
#n,k,x=map(int,input().split())
def countArray(n, k, x):
    # Return the number of ways to fill in the array.
    dp=[[0]*(k+1) for i in range(n+1)]
    dp[1][1]=1
    for i in range(2,n+1):
        dp[i][1]=(dp[i-1][2]*(k-1))%1000000007
        dp[i][2]=(dp[i-1][1] + (dp[i-1][2]*(k-2)))%1000000007
    if(x==1):
        return dp[n][1]
    else:
        return dp[n][2]

print(countArray(n,k,x))
'''