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: ###########
  ###########
  #pythonnnnnnnnnnnnnnnnnnn
  from collections import Counter as co, defaultdict as df
  from copy import deepcopy
  
  def good(frl, k, n):
      
      ans = [0]*n 
      i = 0
      valid = True
      ln = len(frl)
      
      tmp = [ ]
      for i in frl:
          i = i[1]
          if len(i)>=k:
              for j in range(k):
                  ans[i[j]] = j+1 
          else:
              tmp.extend(i)
          
      times = len(tmp)//k 
      while times:
          for i in range(k):
              ans[tmp.pop()] = i+1 
          times -= 1
      
      # while i<ln and valid:
          
      #     kk = k 
      #     if n<kk:
      #         break
          
      #     for j in frl[i][1]:
      #         ans[j] = kk
      #         kk -= 1 
      #         n -= 1
      #         if kk == 0:
      #             break
              
      #     while kk>0:
              
      #         for j in frl[ln-1][1]:
      #             ans[j] = kk
      #             kk -= 1 
      #             n -= 1
      #             if kk == 0:
      #                 break
      #         ln-=1
              
          # i+=1
                  
      return ans, valid
                  
  for _ in range(int(input())):
      
      n,k = map(int, input().split())
      l = list(map(int, input().split()))
      
      uq = df(list)
      
      for i in range(len(l)):
          uq[l[i]].append(i)
      
      frl = sorted([[x, y] for x,y in uq.items()], key = lambda x: len(x[1]), reverse = 1)
      
      ans = [0]*n
      zc = float('inf')
      
      # low = 0
      # up = k 
      # while low<=up:
          
      #     m = (low+up)//2
          
      #     anss, v = good(frl, m, n)
          
      #     tzc = anss.count(0)
          
      #     if v and tzc< zc:
      #         ans = list(anss)
      #         zc = tzc
      #         low = m+1 
      #     else:
      #         up = m-1
      ans , __ = good(frl, k, n)
              
      print(*ans)
          
          
      
      