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 itertools import product, permutations, combinations, accumulate, compress
  from collections import Counter, defaultdict, deque # deque([iterable[, maxlen]]) #appendleft popleft rotate
  from heapq import heapify, heappop, heappush # func(heapifiedlist, item)
  #from bisect import bisect_left, bisect_right, insort # func(sortedlist, item)
  # TODO: more coping mechanisms...?
  #from sys import setrecursionlimit
  
  import os, sys
  input = lambda: sys.stdin.buffer.readline().rstrip(b'\r\n')
  
  def dbp(*args, **kwargs): # calling with dbp(locals()) is perfectly cromulent
  	print(*args, file=sys.stderr, **kwargs)
  
  a_str = lambda: input().decode('ascii')
  get_int_list = lambda: [int(x) for x in input().split()]
  join_str = lambda l, ch=' ': ch.join((str(x) for x in l))  # ch is a default, not kwarg
  
  def IR_SO_DUMN(s, k=2):
  	# bad days happen, I guess
  	d = defaultdict(list)
  	for idx, ch in enumerate(s):
  		d[ch].append(idx)
  
  	ans = [0] * len(s)
  	ccol = best = 0
  	for ch, il in d.items():
  		used = set()
  		for idx in il:
  			used.add(ccol)
  			ans[idx] = ccol+1
  			ccol = ccol+1 # DEAAAAAaaath...
  			if ccol == k:
  				best += 1
  				ccol = 0
  			if ccol in used:
  				break
  	#dbp('case:', s)
  	#dbp(join_str(ans))
  	#best = min((ans.count(c) for c in range(1, k+1))) # lol
  	#dbp('best:', best)
  	c = Counter()
  	for idx, a in enumerate(ans):
  		c[a] += 1
  		if c[a] > best:
  			ans[idx] = 0
  
  
  	return join_str(ans)
  
  
  
  def do_thing():
  	n, k = get_int_list()
  	s = get_int_list()
  	#dbp(s)
  	#war = wa(s)
  	return IR_SO_DUMN(s, k)
  
  def multicase():
  	maxcc = int(input().strip())
  	for cc in range(maxcc):
  		print(do_thing())
  
  if __name__ == "__main__":
  	multicase()
  	#print(do_thing())