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 bisect
  import math
  import collections
  import sys
  import copy
  from functools import reduce
  import decimal
  
  sys.setrecursionlimit(10 ** 9)
  decimal.getcontext().rounding = decimal.ROUND_HALF_UP
  
  graphDict = collections.defaultdict
  queue = collections.deque
  
  
  class Graphs:
      def __init__(self):
          self.graph = graphDict(list)
  
      def add_edge(self, u, v):
          self.graph[u].append(v)
  
      def dfs_utility(self, nodes, visited_nodes):
          visited_nodes.add(nodes)
          for neighbour in self.graph[nodes]:
              if neighbour not in visited_nodes:
                  self.dfs_utility(neighbour, visited_nodes)
              else:
                  return neighbour
  
      def dfs(self, node):
          Visited = set()
          self.dfs_utility(node, Visited)
  
      def bfs(self, node):
          visited = set()
          if node not in visited:
              queue.append(node)
              visited.add(node)
          while queue:
              parent = queue.popleft()
              print(parent)
              for item in self.graph[parent]:
                  if item not in visited:
                      queue.append(item)
                      visited.add(item)
  
  
  def rounding(n):
      return int(decimal.Decimal(f'{n}').to_integral_value())
  
  
  def factors(n):
      return set(reduce(list.__add__,
                        ([i, n // i] for i in range(1, int(n ** 0.5) + 1) if n % i == 0)))
  
  
  def inp():
      return sys.stdin.readline().strip()
  
  
  def map_inp(v_type):
      return map(v_type, inp().split())
  
  
  def list_inp(v_type):
      return list(map_inp(v_type))
  
  
  ######################################## Solution ####################################
  
  for _ in range(int(inp())):
      n, k = map_inp(int)
      s = list_inp(int)
      dic = collections.Counter(s)
      count = 0
      ans = 0
      index__ = {}
      for i in range(len(s)):
          if s[i] not in index__:
              index__[s[i]] = []
              index__[s[i]].append(i)
          else:
              index__[s[i]].append(i)
  
      index = []
      for item in dic.keys():
          if dic[item] >= k:
              count += k
              temp = 0
              for i in index__[item]:
                  if s[i] == item and temp < k:
                      index.append(i)
                      temp += 1
          else:
              count += dic[item]
              for i in index__[item]:
                  if s[i] == item:
                      index.append(i)
      count -= count % k
      i = 1
      ans = [0 for item in range(n)]
      for item in index[:count]:
          if i <= k:
              ans[item] = i
              i += 1
          else:
              i = 1
              ans[item] = i
              i += 1
      print(*ans)
  