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 sys
  from sys import stdout
  
  input = lambda: sys.stdin.readline().strip()
  P = lambda: list(map(int, input().split()))
  from math import factorial as f, gcd
  from collections import deque, defaultdict as dd, Counter as C
  from heapq import heapify, heappop, heappush, heappushpop, heapreplace, merge
  from random import randint, choice, sample
  import time
  mod = 10**9+7
  a = ord('a')
  
  # Fenwick Tree
  class FT:
      def __init__(self, N):
          self.ft = [0] * (N+1)
          self.N = N
  
      def ls(self, x):
          return x & -x
  
      def fenwick_update(self, p, v):
          while p <= self.N:
              self.ft[p] += v
              p += self.ls(p)
  
      def fenwick_query(self, p):
          s = 0
          while p:
              s += self.ft[p]
              p -= self.ls(p)
          # print(p, s)
          return s
  
  start = time.time()
  def fast_exp(x, exp):
      ans = 1
      base = x
      while exp:
          if exp & 1:
              ans *= base
          base *= base
          base %= mod
          ans %= mod
          exp >>= 1
      return ans
  
  def countBits(n):
      count = 0
      while n:
          count += n & 1
          n >>= 1
      return count
  
  def submasks(n):
      #this is cool
      #https://cp-algorithms.com/algebra/all-submasks.html
      org = n
      while n:
          yield n
          n = (n-1) & org
  
  
  def solve():
      n, k = P()
      L = P()
      c = list(C(L).items())
      c.sort(key=lambda i:-i[1])
      c2 = dd(list)
      for i, x in enumerate(L):
          c2[x].append(i)
  
      # print(c)
      # less = left = 0
      ans = [0] * (n)
      curr = 1
      s = []
      co = 0
      for kk, v in c:
          v = min(v, k)
          for i in range(v):
              topaint = c2[kk][i]
              ans[topaint] = curr
              curr += 1
              if curr == k+1:
                  curr = 1
              s.append(topaint)
              co += 1
      for _ in range(co % k):
          ans[s.pop()] = 0
      print(*ans)
  
  
  
  
      
  
  
  
  
  
  
  tc = int(input())
  for t in range(1, tc+1):
      solve()
  
  # solve()
  # print(time.time()-start)
  
  
  