description: This problem is an extension of the problem "Wonderful Coloring - 1". It has quite many differences, so you should read this statement completely.
  
  Recently, Paul and Mary have found a new favorite sequence of integers a_1, a_2, ..., a_n. They want to paint it using pieces of chalk of k colors. The coloring of a sequence is called wonderful if the following conditions are met:
  
    1. each element of the sequence is either painted in one of k colors or isn't painted; 
    2. each two elements which are painted in the same color are different (i. e. there's no two equal values painted in the same color); 
    3. let's calculate for each of k colors the number of elements painted in the color — all calculated numbers must be equal; 
    4. the total number of painted elements of the sequence is the maximum among all colorings of the sequence which meet the first three conditions. 
  
  
  
  E. g. consider a sequence a=[3, 1, 1, 1, 1, 10, 3, 10, 10, 2] and k=3. One of the wonderful colorings of the sequence is shown in the figure.
  
  <image> The example of a wonderful coloring of the sequence a=[3, 1, 1, 1, 1, 10, 3, 10, 10, 2] and k=3. Note that one of the elements isn't painted.
  
  Help Paul and Mary to find a wonderful coloring of a given sequence a.
  
  Input
  
  The first line contains one integer t (1 ≤ t ≤ 10000) — the number of test cases. Then t test cases follow.
  
  Each test case consists of two lines. The first one contains two integers n and k (1 ≤ n ≤ 2⋅10^5, 1 ≤ k ≤ n) — the length of a given sequence and the number of colors, respectively. The second one contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n).
  
  It is guaranteed that the sum of n over all test cases doesn't exceed 2 ⋅ 10^5.
  
  Output
  
  Output t lines, each of them must contain a description of a wonderful coloring for the corresponding test case.
  
  Each wonderful coloring must be printed as a sequence of n integers c_1, c_2, ..., c_n (0 ≤ c_i ≤ k) separated by spaces where
  
    * c_i=0, if i-th element isn't painted; 
    * c_i>0, if i-th element is painted in the c_i-th color. 
  
  
  
  Remember that you need to maximize the total count of painted elements for the wonderful coloring. If there are multiple solutions, print any one.
  
  Example
  
  Input
  
  
  6
  10 3
  3 1 1 1 1 10 3 10 10 2
  4 4
  1 1 1 1
  1 1
  1
  13 1
  3 1 4 1 5 9 2 6 5 3 5 8 9
  13 2
  3 1 4 1 5 9 2 6 5 3 5 8 9
  13 3
  3 1 4 1 5 9 2 6 5 3 5 8 9
  
  
  Output
  
  
  1 1 0 2 3 2 2 1 3 3
  4 2 1 3
  1
  0 0 1 1 0 1 1 1 0 1 1 1 0
  2 1 2 2 1 1 1 1 2 1 0 2 2
  1 1 3 2 1 3 3 1 2 2 3 2 0
  
  Note
  
  In the first test case, the answer is shown in the figure in the statement. The red color has number 1, the blue color — 2, the green — 3.
solution: from __future__ import division, print_function
  
  import os,sys
  from io import BytesIO, IOBase
  
  if sys.version_info[0] < 3:
      from __builtin__ import xrange as range
      from future_builtins import ascii, filter, hex, map, oct, zip
  
  from bisect import bisect_left as lower_bound, bisect_right as upper_bound 
  def so():  return int(input())
  def st():  return input()
  def mj():  return map(int,input().strip().split(" "))
  def msj(): return map(str,input().strip().split(" "))
  def le():  return list(map(int,input().split()))
  def lebe():return list(map(int, input()))
  
  def dmain():
      sys.setrecursionlimit(1000000)
      threading.stack_size(1024000)
      thread = threading.Thread(target=main)
      thread.start()
  def joro(L):
      return(''.join(map(str, L)))
  
  
  def decimalToBinary(n): return bin(n).replace("0b","")
  
  
  def isprime(n):
      for i in range(2,int(n**0.5)+1):
          if n%i==0:
              return False
      return True
          
      
  def read():
      sys.stdin  = open('input.txt', 'r')  
      sys.stdout = open('output.txt', 'w') 
  def tr(n):
      return n*(n+1)//2
  
  
  
  
  def main():
      for i in range(so()):
          n,q=mj()
          L=le()
          B=[]
          d=0
          for i in range(n):
              B.append(0)
          Z=[[]for i in range(1+n)]
          M=[]
          P=[]
          Y=[]
          for i in range(1+n):
              M.append(0)
          for i in range(1+n):
              P.append(0)
          for u in range(n):
              if q>P[L[u]]:
                  P[L[u]]+=1
                  Z[L[u]].append(u)
          for r in range(1+n):
              for v in Z[r]:
                  Y.append(v)
          
          er=(len(Y)//q)*q
          for p in range(0,er,q):
              bec=0
              for ut in range(p,q+p,1):
                  B[Y[ut]]=1+bec
                  bec=1+bec
          for i in B:
              print(i,end=" ")
          print('\n')
                  
                  
          
                  
          
                  
          
          
          
          
              
          
          
              
              
              
          
      
          
          
      
      
      
      
          
          
      
              
          
     
         
      
             
            
            
                  
              
          
                  
          
          
         
              
                  
          
  
  
  
  
  # region fastio
  # template taken from https://github.com/cheran-senthil/PyRival/blob/master/templates/template.py
  
  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")
  
  # endregion
  
  
  if __name__ == "__main__":
      #read()
      main()
      #dmain()
  
  # Comment Read()
  
  
  
  
  
      
  
          
  
  
  
  
  