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: #----------FASTIOSTART-----------#
  from __future__ import division, print_function
  import os
  import 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
  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):
  	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")
  #----------FASTIOFINISH----------#
  import collections,string,bisect,re,random,queue,itertools,statistics,math
  from collections import *
  from bisect import *
  from string import *
  from itertools import * 
  from statistics import *
  from math import *
  from re import *
  from queue import *
  #----------SASTA_STL-------------#
  # stack
  class Stack:
      def __init__(self):self.items = []
      def push(self, item):self.items.append(item)
      def pop(self):return self.items.pop()
      def empty(self):return self.items == []
      def size(self):return len(self.items)
      def at(self,idx):
      	try:return self.items[idx]
      	except:return -1
  #priority_queue
  class priority_queue(object):
      def __init__(self):self.queue = []
      def __str__(self):return ' '.join([str(i) for i in self.queue])
      def empty(self):return len(self.queue) == 0
      def push(self, data):self.queue.append(data)
      def pop(self):
          try:
              max = 0
              for i in range(len(self.queue)):
                  if self.queue[i] > self.queue[max]:
                      max = i
              item = self.queue[max]
              del self.queue[max]
              return item
          except IndexError:
              print()
              exit()
  #----------SASTA_STL-------------#
  mod  = int(1e9+7)
  imax = float("inf")
  imin = float("-inf")
  true = True
  false= False
  N 	 = int(1e5+10)
  none = None
  inp = lambda : input()
  I   = lambda : int(inp())
  M   = lambda : map(int,inp().split())
  MS  = lambda : map(str,inp().split())
  S   = lambda : list(MS())
  L   = lambda : list(M())
  def IO():
  	try:sys.stdin = open('input.txt', 'r');sys.stdout = open('uttar.txt', 'w')
  	except:pass
  IO()
  #----------TOTKA---------#
  
  def kabraji_ka_totka():
  	# totka hai dosto !
  	n,k=L()
  	arr=L()
  	cnt=[0]*(n+1)
  	ind=[]
  	for i,j in enumerate(arr):
  		if cnt[j] < k:
  			ind.append((j,i))
  		cnt[j]+=1
  	ind.sort()
  	res=0
  
  	m=[0]*n;
  	m1=len(ind)-len(ind)%k;
  	for x,i in ind[:m1]:
  		m[i]=res+1;
  		res=(res+1)%k;
  	print(*m)
  
  #----------TOTKA----------#
  if __name__ == '__main__':
  	for i in range(I()):kabraji_ka_totka()