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:
import sys
import bisect as bi
import math
from collections import defaultdict as dd
input=sys.stdin.readline
##sys.setrecursionlimit(10**7)
def cin():
    return map(int,sin().split())
def ain():            
    return list(map(int,sin().split()))
def sin():
    return input()
def inin():
    return int(input())
for _ in range(inin()):
    n=inin()
    l=ain()
    s=set();ans=[]
    for i in l:
        if i not in s:
            s.add(i)
            ans+=[i]
    print(*ans)
    
        

    
    
    
            
    

    
####    s=input()
####    def minCut( s: str) -> int:
####    
####        n=len(s)
####        dp=[[-1]*(n+2) for i in range(n+2)]
####        def pp(s,i,j,dp,mn):
####            if(i>=j or s[i:j+1]==s[i:j+1:-1]):return 0
####            if(dp[i][j]!=-1):return dp[i][j]
####            for k in range(i,j):
####                if(dp[i][k]!=-1):return dp[i][k]
####                else:
####                    dp[i][k]=pp(s,i,k,dp,mn)
####                    return dp[i][k]
####                if(dp[k+1][j]!=-1):return dp[k+1][j]
####                else:
####                    dp[i][k]=pp(s,k+1,j,dp,mn)
####                    return dp[i][k]
####                temp=dp[i][k] + dp[i][k] +1
####                if(temp<mn):mn=tmp
####            dp[0][n]=mn
####            return dp[0][n]
####        x=pp(s,0,n,dp,999999999999999)
####        for i in dp:print(i)
####        return x
####    
####    print(minCut(s))
##dp=[[0 if (i==j) else -1 for j in range(n+1)] for i in range(n+1)]
##for i in range(1,n):
##    for j in range(
##
##
##
##
##        
##    
##    
####n=m=0
####s=''
####t=''
####dp=[]
####def solve(inds,indt,k,cont):
####    ans=-999999999999999
####    print(dp)
####    if(k<0):return 0
####    elif(inds>=n and indt>=m):return 0
####    elif(dp[inds][indt][k][cont]!=-1):return dp[inds][indt][k][cont]
####    else:
####        if(indt<m):ans=max(ans,solve(inds,indt+1,k,0))
####        if(inds<n):ans=max(ans,solve(inds+1,indt,k,0))
####        if(s[inds]==t[indt]):
####            ans=max(ans,solve(inds+1,indt+1,k-1,1)+1)
####            if(cont):ans=max(ans,solve(inds+1,indt+1,k,1)+1)
####        dp[inds][indt][k][cont]=ans
####        return ans    
##
####    n,m,k=cin()
####    s=sin().strip()
####    t=sin().strip()
####    dp=[[[[-1]*2 for i in range(k)] for i in range(m+1)] for i in range(n+1)]
####    c=0
####    for i in dp:
####        for j in i:
####            for l in j:
####                c+=1
####                print(l,c)
####    print(solve(0,0,k,0))