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
  from math import ceil, gcd, sqrt
  import string
  from decimal import Decimal
  
  
  def binary_table(string_with_all_characters, length_to_make):
      return [''.join(x) for x in product(string_with_all_characters, repeat=length_to_make)]
  
  
  def all_possible_substrings(string):
      return [int(string[i: j]) for i in range(len(string)) for j in range(i + 1, len(string) + 1)]
  
  
  def number_of_substrings(length):
      return int(length * (length + 1) / 2)
  
  
  for enumeration in range(int(input())):
      temp = {}
      num_of_elements, num_of_colors = map(int, input().split())
      array = list(map(int, input().split()))
      current_color = 1
      indexes = {}
  
      for i in range(len(array)):
          value = array[i]
          if temp.get(value) is None:
              temp[value] = 1
          else:
              temp[value] += 1
  
          if indexes.get(value) is None:
              indexes[value] = [i]
          else:
              indexes[value].append(i)
  
          if temp[value] > num_of_colors:
              array[i] = 0
              indexes[value].remove(i)
              temp[value] -= 1
      length_value = len([x for x in array if x > 0])
      num = 0
      #print(indexes)
      while length_value % num_of_colors:
          if array[num] != 0:
              indexes[array[num]].remove(num)
              array[num] = 0
              length_value -= 1
          num += 1
      #print(temp)
      #print(array)
      build = array[::]
      #print(indexes)
      for i in temp:
          possible_index = indexes[i]
          #print(i, possible_index)
          for j in possible_index:
              #print(j, current_color)
              build[j] = current_color
              current_color += 1
              if current_color > num_of_colors:
                  current_color = 1
      #print(array)
      print(*build)
  