human
Write me a Python code to print the answer: n integers p_1, p_2, given the following:
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 
Using python3 can you solve the prior task?
assistant
#Here is Python code to print the answer: n integers p_1, p_2, .

#!/bin/python3

import math
import os
import random
import re
import sys
from collections import defaultdict

# # def getMapCount(mapping, key):


# #     if key in mapping:
# #         return mapping[key]
# #     else:
# #         return 0

# def binSearch(arr, item):
#     low, hi = 0, len(arr) - 1
#     while low <= hi:
#         mid = (low + hi) // 2

#         if arr[mid] < item:
#             low = mid + 1
#         else:
#             hi = mid -1 
#             ans = mid
#         mid = (low + hi )// 2
    
#     return mid

# #exact binarySearch
# def binarySearch (arr, item): 
#     low, hi = 0, len(arr) - 1
    
#     while low <= hi:
#         mid = (low + hi) // 2

#         if arr[mid] == x: 
#             return mid 
        
#          if arr[mid] < item:
#             low = mid + 1
#         else:
#             hi = mid -1 
#     return -1

  

# # print(binSearch([1,2,4,5,9,11,13,99,100], 8))
# def countTriplets(arr, r):
#     # arr = sorted(arr)
#     countMap = dict()
#     index = 0
#     for item in arr:
#         if item not in countMap:
#             countMap[item] = [index]

#         else:
#             countMap[item].append(index)

#         index += 1

#     count = 0
#     print(countMap)
#     indx  = 0
#     for key in countMap:
#         for indx in countMap[key]:
#             if r == 1:
#                 count += (countMap[key] * (countMap[key] - 1) * (countMap[key] - 2)) // 6
#             else:
#                 count += countMap[key] * getMapCount(countMap, indx, key * r) * getMapCount(countMap, key * r * r)
#         indx += 1

#     return count

# fib_ara = [1,1]
# def fib():
#     while fib_ara[-1] < 10**9:
#         fib_ara.append(fib_ara[-1] + fib_ara[-2])

# fib()

case = int(input())
 
for i in range(case):
    # a = [int(x) for x in  input().split() ]
    # x,y,z = a[0],a[1], a[2]
    n = int(input())
    ara = [int(x) for x in  input().split() ]
    hasTaken = {}
    perm = []
    for item in ara:
        if item not in hasTaken:
            perm.append(item)
...