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 os.path
  from math import gcd, floor, ceil
  from collections import *
  import sys
  mod = 1000000007
  INF = float('inf')
  def st(): return list(sys.stdin.readline().strip())
  def li(): return list(map(int, sys.stdin.readline().split()))
  def mp(): return map(int, sys.stdin.readline().split())
  def inp(): return int(sys.stdin.readline())
  def pr(n): return sys.stdout.write(str(n)+"\n")
  def prl(n): return sys.stdout.write(str(n)+" ")
  
  
  if os.path.exists('input.txt'):
      sys.stdin = open('input.txt', 'r')
      sys.stdout = open('output.txt', 'w')
  
  # joined late
  
  
  def solve():
      n, k = mp()
      l = li()
      d = Counter(l)
      ind = defaultdict(list)
      for i, ele in enumerate(l):
          ind[ele].append(i)
      found = set()
      odds = 0
      not_found = set()
      for i in d:
          if d[i] >= k:
              found.add(i)
          else:
              not_found.add(i)
              odds += d[i]
      ans = [0]*n
      for i in found:
          count = 0
          for j in ind[i]:
              ans[j] = count + 1
              count += 1
              if count == k:
                  break
      round = 0
      count = 0
      for i in not_found:
          for j in ind[i]:
              if count == k:
                  count = 0
                  round += 1
              if round >= (odds//k):
                  break
              ans[j] = count + 1
              count += 1
      # print(Counter(ans))
      print(*ans)
  
  
  for _ in range(inp()):
      solve()
  