description: You are given an array a of n integers, and another integer k such that 2k ≤ n.
  
  You have to perform exactly k operations with this array. In one operation, you have to choose two elements of the array (let them be a_i and a_j; they can be equal or different, but their positions in the array must not be the same), remove them from the array, and add ⌊ (a_i)/(a_j) ⌋ to your score, where ⌊ x/y ⌋ is the maximum integer not exceeding x/y.
  
  Initially, your score is 0. After you perform exactly k operations, you add all the remaining elements of the array to the score.
  
  Calculate the minimum possible score you can get.
  
  Input
  
  The first line of the input contains one integer t (1 ≤ t ≤ 500) — the number of test cases.
  
  Each test case consists of two lines. The first line contains two integers n and k (1 ≤ n ≤ 100; 0 ≤ k ≤ ⌊ n/2 ⌋).
  
  The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 2 ⋅ 10^5).
  
  Output
  
  Print one integer — the minimum possible score you can get.
  
  Example
  
  Input
  
  
  5
  7 3
  1 1 1 2 1 3 1
  5 1
  5 5 5 5 5
  4 2
  1 3 3 7
  2 0
  4 2
  9 2
  1 10 10 1 10 2 7 10 3
  
  
  Output
  
  
  2
  16
  0
  6
  16
  
  Note
  
  Let's consider the example test.
  
  In the first test case, one way to obtain a score of 2 is the following one:
  
    1. choose a_7 = 1 and a_4 = 2 for the operation; the score becomes 0 + ⌊ 1/2 ⌋ = 0, the array becomes [1, 1, 1, 1, 3]; 
    2. choose a_1 = 1 and a_5 = 3 for the operation; the score becomes 0 + ⌊ 1/3 ⌋ = 0, the array becomes [1, 1, 1]; 
    3. choose a_1 = 1 and a_2 = 1 for the operation; the score becomes 0 + ⌊ 1/1 ⌋ = 1, the array becomes [1]; 
    4. add the remaining element 1 to the score, so the resulting score is 2. 
  
  
  
  In the second test case, no matter which operations you choose, the resulting score is 16.
  
  In the third test case, one way to obtain a score of 0 is the following one:
  
    1. choose a_1 = 1 and a_2 = 3 for the operation; the score becomes 0 + ⌊ 1/3 ⌋ = 0, the array becomes [3, 7]; 
    2. choose a_1 = 3 and a_2 = 7 for the operation; the score becomes 0 + ⌊ 3/7 ⌋ = 0, the array becomes empty; 
    3. the array is empty, so the score doesn't change anymore. 
  
  
  
  In the fourth test case, no operations can be performed, so the score is the sum of the elements of the array: 4 + 2 = 6.
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,k=di()
      a=sorted(li(), reverse=True)
      if k==0:
          print(sum(a))
      else:
          score=0
          for i in range(k):
              temp=a[i+k]//a[i]
              score+=temp
          score+=sum(a[2*k:])
          print(score)
              
  
  
          