description: You are given a sequence a_1, a_2, ..., a_n consisting of n pairwise distinct positive integers.
  
  Find \left⌊ \frac n 2 \right⌋ different pairs of integers x and y such that: 
  
    * x ≠ y; 
    * x and y appear in a; 
    * x~mod~y doesn't appear in a. 
  
  
  
  Note that some x or y can belong to multiple pairs.
  
  ⌊ x ⌋ denotes the floor function — the largest integer less than or equal to x. x~mod~y denotes the remainder from dividing x by y.
  
  If there are multiple solutions, print any of them. It can be shown that at least one solution always exists.
  
  Input
  
  The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of testcases.
  
  The first line of each testcase contains a single integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the length of the sequence.
  
  The second line of each testcase contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^6).
  
  All numbers in the sequence are pairwise distinct. The sum of n over all testcases doesn't exceed 2 ⋅ 10^5.
  
  Output
  
  The answer for each testcase should contain \left⌊ \frac n 2 \right⌋ different pairs of integers x and y such that x ≠ y, x and y appear in a and x~mod~y doesn't appear in a. Print the pairs one after another.
  
  You can print the pairs in any order. However, the order of numbers in the pair should be exactly such that the first number is x and the second number is y. All pairs should be pairwise distinct.
  
  If there are multiple solutions, print any of them.
  
  Example
  
  Input
  
  
  4
  2
  1 4
  4
  2 8 3 4
  5
  3 8 5 9 7
  6
  2 7 5 3 4 8
  
  
  Output
  
  
  4 1
  8 2
  8 4
  9 5
  7 5
  8 7
  4 3
  5 2
  
  Note
  
  In the first testcase there are only two pairs: (1, 4) and (4, 1). \left⌊ \frac 2 2 \right⌋=1, so we have to find one pair. 1~mod~4=1, and 1 appears in a, so that pair is invalid. Thus, the only possible answer is a pair (4, 1).
  
  In the second testcase, we chose pairs 8~mod~2=0 and 8~mod~4=0. 0 doesn't appear in a, so that answer is valid. There are multiple possible answers for that testcase.
  
  In the third testcase, the chosen pairs are 9~mod~5=4 and 7~mod~5=2. Neither 4, nor 2, appears in a, so that answer is valid.
solution: #Mamma don't raises quitter.................................................
  #manan is improving, keep hustling manan bhai ;)
  from collections import deque as de
  import math
  import sys
  from math import sqrt as sq
  from math import floor as fl
  from math import ceil as ce
  from sys import stdin, stdout
  import time
  
  sys.setrecursionlimit(10**7)
  from collections import Counter as cnt
  from functools import  reduce
  
  from itertools import groupby as gb
  #from fractions import Fraction as fr
  from bisect import bisect_left as bl, bisect_right as br
  from typing import Coroutine
  
  def factors(n):    
      return set(reduce(list.__add__, 
                  ([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0)))
      
  class My_stack():
      def __init__(self):
          self.data = []
      def my_push(self, x):
          return (self.data.append(x))
      def my_pop(self):
          return (self.data.pop())
      def __iter__(self):
          return iter(self.data)
      def my_peak(self):
          return (self.data[-1])
      def my_contains(self, x):
          return (self.data.count(x))
      def my_show_all(self):
          return (self.data)
      def isEmpty(self):
        return len(self.data)==0
      def len(self):
          return len(self.data)
      
  
  arrStack = My_stack() 
  
  class graph():
      def __init__(self):
          self.graphdict = {}
      def add_edge(self, node, neighbor):
          if node not in self.graphdict:
              self.graphdict[node]=[neighbor]
          else:
              self.graphdict[node].append(neighbor)
      def show_edge(self):
          for node in self.graphdict:
              for neighbors in self.graphdict:
                  print(node, neighbors)
      def show_graph(self):
          return self.graphdict
  
      
  
  
      
              
  
  #decimal to binary   
  def decimalToBinary(n): 
      return bin(n).replace("0b", "")
  #binary to decimal
  def binarytodecimal(n):
      return int(n,2)
  
  def isPrime(n) : 
  	if (n <= 1) : 
  		return False
  	if (n <= 3) : 
  		return True
  	if (n % 2 == 0 or n % 3 == 0) : 
  		return False
  	i = 5
  	while(i * i <= n) : 
  		if (n % i == 0 or n % (i + 2) == 0) : 
  			return False
  		i = i + 6
  	return True
  
  def get_prime_factors(number):
      prime_factors = []
      while number % 2 == 0:
          prime_factors.append(2)
          number = number / 2
      for i in range(3, int(math.sqrt(number)) + 1, 2):
          while number % i == 0:
              prime_factors.append(int(i))
              number = number / i
  
      if number > 2:
          prime_factors.append(int(number))
      return prime_factors
  
  def get_frequency(list):
      dic={}
      for ele in list:
          if ele in dic:
              dic[ele] += 1
          else:
              dic[ele] = 1
      return dic
  
  def Log2(x): 
      return (math.log10(x) / 
              math.log10(2)); 
  
   
  # Function to get product of digits
  def getProduct(n):
      product = 1
      while (n != 0):
          product = product * (n % 10)
          n = n // 10
      return product
  #fucntion to find regular bracket sequence
  def regularBracketsequence(str,arrStack):
      for i in range(len(str)):
          if str[i] =='(':
              arrStack.my_push(str[i])
          else:
              if arrStack.isEmpty():
                  return False
              else:
                  arrStack.my_pop()
      if arrStack.isEmpty():
          return True
      else:
          return False
  
  #function to find LCM of two numbers
  def lcm(x,y):
     lcm = (x*y)//math.gcd(x,y)
     return lcm
  
  def isPowerOfTwo(n): 
      return (math.ceil(Log2(n)) == math.floor(Log2(n))); 
  #to check whether the given sorted sequnce is forming an AP or not....
  def checkisap(list):
      d=list[1]-list[0]
      for i in range(2,len(list)):
          temp=list[i]-list[i-1]
          if temp !=d:
              return False
      return True
  
  #seive of erathanos
  def primes_method5(n):
      out ={}
      sieve = [True] * (n+1)
      for p in range(2, n+1):
          if (sieve[p]):
              out[p]=1
              for i in range(p, n+1, p):
                  sieve[i] = False
      return out
  #fucntion to get unique number to a cell in a matrix
  def id(i, j, col): #one based indexing and columnwise numeration
      ans= (i-1)*col
      ans+=j
      return ans
  
  #function to get the sum of digits
  def getSum(n): 
        
      strr = str(n)
      list_of_number = list(map(int, strr.strip()))
      return sum(list_of_number)
  def checkperfectsquare(n):
      if (ce(sq(n)) == fl(sq(n))):
          return True
      else:
          return False
  #this used for selecting r objects from n ojects
  def combinatrics(n,r):
      num=n
      deno=n-r
      pr=1
      for i in range(num,deno,-1):
          pr*=i
      an=1
      for i in range(1,r+1):
          an*=i
      ans=pr//an
      return ans
  #ceil  function gives wrong answer after 10^17 so i have to create my own :)
  # because i don't want to doubt on my solution of 900-1000 problem set.
  def ceildiv(x,y): 
      return (x+y-1)//y 
    
  def di():return map(int, input().split())
  def ii():return int(input())
  def li():return list(map(int, input().split()))
  def si():return list(map(str, input()))
  def indict():
      dic = {}
      for index, value in enumerate(input().split()):
          dic[int(value)] = int(index)
      return dic
  def frqdict(): 
      # by default it is for integer input. :)
      dic={}
      for index, value in enumerate(input()):
          if value not in dic:
              dic[value] =1
          else:
              dic[value] +=1
      return dic
  
  #sys.stdin = open('input.txt', 'r')
  #sys.stdout = open('output.txt', 'w')
  #practice like your never won
  #perform like you never lost
  t=ii()
  while t:
      t-=1
      n=ii()
      a=sorted(li())
      for i in range(1,(n//2)+1):
          print( a[i], a[0])
      