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: import functools
  import itertools
  import math
  import random
  import string
  from collections import defaultdict, Counter
  from heapq import *
  from bisect import bisect_left, bisect_right
  
  
  def repeat(_func=None, *, times=1):
      def decorator(func):
          @functools.wraps(func)
          def wrapper(*args, **kwargs):
              for _ in range(times):
                  func(*args, **kwargs)
  
          return wrapper
  
      if _func is None:
          return decorator
      else:
          return decorator(_func)
  
  
  def unpack(func=int):
      return map(func, input().split())
  
  
  def l_unpack(func=int):
      """list unpack"""
      return list(map(func, input().split()))
  
  
  def getint():
      return int(input())
  
  
  def getmatrix(rows):
      return [list(map(int, input().split())) for _ in range(rows)]
  
  
  def display_matrix(mat):
      for i in range(len(mat)):
          print(mat[i])
      print('-' * 20)
  
  
  MOD = 1_000_000_007
  
  
  @repeat(times=int(input()))
  def main():
      def chain():
          for v in hmap.values():
              yield from v
  
      n, k = unpack()
      arr = l_unpack()
      out = [0] * n
      if k == 0:
          print(*out)
          return
      total = 0
      hmap = defaultdict(list)
      for i, c in enumerate(arr):
          if len(hmap[c]) < k:
              hmap[c].append(i)
              total += 1
      each = total // k
      color = [(0, i) for i in range(1, k + 1)]
      for pos in chain():
          while color and color[0][0] == each:
              heappop(color)
          if not total or not color:
              break
          val, col = color[0]
          out[pos] = col
          total -= 1
          heapreplace(color, (val + 1, col))
      print(*out)
  
  
  main()
  