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: from __future__ import division, print_function
  import math
  import sys
  import os
  from io import BytesIO, IOBase
  #from collections import deque, Counter, OrderedDict, defaultdict
  #import heapq
  #ceil,floor,log,sqrt,factorial,pow,pi,gcd
  #import bisect
  #from bisect import bisect_left,bisect_right
  
  BUFSIZE = 8192
  
  
  class FastIO(IOBase):
  	newlines = 0
  
  	def __init__(self, file):
  		self._fd = file.fileno()
  		self.buffer = BytesIO()
  		self.writable = "x" in file.mode or "r" not in file.mode
  		self.write = self.buffer.write if self.writable else None
  
  	def read(self):
  		while True:
  			b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
  			if not b:
  				break
  			ptr = self.buffer.tell()
  			self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
  		self.newlines = 0
  		return self.buffer.read()
  
  	def readline(self):
  		while self.newlines == 0:
  			b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
  			self.newlines = b.count(b"\n") + (not b)
  			ptr = self.buffer.tell()
  			self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
  		self.newlines -= 1
  		return self.buffer.readline()
  
  	def flush(self):
  		if self.writable:
  			os.write(self._fd, self.buffer.getvalue())
  			self.buffer.truncate(0), self.buffer.seek(0)
  
  
  class IOWrapper(IOBase):
  	def __init__(self, file):
  		self.buffer = FastIO(file)
  		self.flush = self.buffer.flush
  		self.writable = self.buffer.writable
  		self.write = lambda s: self.buffer.write(s.encode("ascii"))
  		self.read = lambda: self.buffer.read().decode("ascii")
  		self.readline = lambda: self.buffer.readline().decode("ascii")
  
  
  def print(*args, **kwargs):
  	"""Prints the values to a stream, or to sys.stdout by default."""
  	sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
  	at_start = True
  	for x in args:
  		if not at_start:
  			file.write(sep)
  		file.write(str(x))
  		at_start = False
  	file.write(kwargs.pop("end", "\n"))
  	if kwargs.pop("flush", False):
  		file.flush()
  
  
  if sys.version_info[0] < 3:
  	sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
  else:
  	sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
  
  input = lambda: sys.stdin.readline().rstrip("\r\n")
  
  def I():
      return(int(input()))
  def lint():
      return(list(map(int,input().split())))
  def insr():
      s = input().strip()
      return(list(s[:len(s)]))
  def invr():
      return(map(int,input().split()))
  import itertools
  from sys import maxsize, stdout, stdin,stderr
  mod = int(1e9+7)
  import sys
  # def I(): return int(stdin.readline())
  # def lint(): return [int(x) for x in stdin.readline().split()]
  def S(): return list(map(str,input().strip()))
  def grid(r, c): return [lint() for i in range(r)]
  from collections import defaultdict, Counter, deque
  import math
  import heapq
  from heapq import heappop , heappush
  import bisect
  from math import factorial, inf
  from itertools import groupby
  from itertools import permutations as comb
  
  mod=1000000007
  MOD = 998244353
  def gcd(a,b): 
      while b:
          a %= b
          tmp = a
          a = b
          b = tmp
      
      return a
   
  def lcm(a,b): 
      return a  // gcd(a, b) * b
   
  def check_prime(n):
      for i in range(2, int(n ** (1 / 2)) + 1):
          if not n % i:
              return False
      return True
  
  # def nCr(n, r):
   
  #     return (fact(n) // (fact(r)
  #                 * fact(n - r)))
   
  # Returns factorial of n
  # def fact(n):
   
  #     res = 1
       
  #     for i in range(2, n+1):
  #         res = res * i
           
  #     return res
  def primefactors(n):
      num=0
      
      while n % 2 == 0:
          num+=1
          n = n / 2
      
      for i in range(3,int(math.sqrt(n))+1,2):
           
      
          while n % i== 0:
              num+=1
              n = n // i
                 
      
      if n > 2:
          num+=1
      return num
  '''
  def iter_ds(src):
      store=[src]
      while len(store):
          tmp=store.pop()
          if not vis[tmp]:
              vis[tmp]=True
              for j in ar[tmp]:
                  store.append(j)
  '''
  def ask(a,b,c):
      # print('? 1 {}'.format(a),flush=True)
      print(c,a,b,flush=True)
      n=I()
      
      return n
  def linear_sieve(n):
      is_composite=[False]*n
      prime=[]
      for i in range(2,n):
          if not is_composite[i]:
              prime.append(i)
          for j in prime:
              is_composite[i*j]=True
              if i%prime==0:
                  break
      return prime
   
  def dfs(i,p,d):
      
      a,tmp=0,0
      for j in d[i]:
          if j!=p:
              a+=1
              tmp+=dfs(j,i)
      
      if a==0:
          return 0
      
      return tmp/a + 1 
  def primeFactors(n):
      l=[]
      
      while n % 2 == 0:
          l.append(2)
          n = n // 2
      for i in range(3,int(math.sqrt(n))+1,2):
           
          
          while n % i== 0:
              l.append(i)
              n = n // i
  
      if n > 2:
          l.append(n)
      return l
  
  def binpow(a,b,c):
      res=1
      a=a%c
      while b>0:
          if b&1:
              res=(res*a)%c
  
          b=b>>1
          a = (a*a)%c
  
      return res
  
  
      
  
  #  Sieve
  d=[]
  primes=[]
  prim=[0]*(10**6+1)
  def sieve(n):
      
      for i in range(n):
          d.append(i)
  
      for i in range(2,n):
  
          if d[i]==i:
              prim[i]=1
              primes.append(i)
          j=0
          while j<len(primes) and primes[j]<=d[i] and primes[j]*i<n:
          
              d[i * primes[j]] = primes[j]
              j+=1
  
          
  def primeFactors(n):
      factors=[]
      while n!=1:
          factors.append(d[n])
          n//=d[n]
      
      return factors
  
  t = 1
  t = I()
  for _ in range(t):
      n = I()
      s = lint()
      d=Counter(s)
      s.sort()
      cnt=0
      for i in range(n):
          for j in range(i+1,n):
              if s[i]!=s[j]:
                  tmp=s[j]%s[i]
                  if d[tmp]==0:
                      print(s[j],s[i])
                      cnt+=1
              if cnt==n//2:
                  break
          if cnt==n//2:
              break
  
  
  
       
  
      
    
  
          
  
  
          
      
   
  
  