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
  import bisect
  from bisect import bisect_left as lb
  from bisect import bisect_right as rb
  input_=lambda: sys.stdin.readline().strip("\r\n")
  from math import log
  from math import gcd
  from math import atan2,acos
  from random import randint
  sa=lambda :input_()
  sb=lambda:int(input_())
  sc=lambda:input_().split()
  sd=lambda:list(map(int,input_().split()))
  sflo=lambda:list(map(float,input_().split()))
  se=lambda:float(input_())
  sf=lambda:list(input_())
  flsh=lambda: sys.stdout.flush()
  #sys.setrecursionlimit(10**6)
  mod=10**9+7
  mod1=998244353
  gp=[]
  cost=[]
  dp=[]
  mx=[]
  ans1=[]
  ans2=[]
  special=[]
  specnode=[]
  a=0
  kthpar=[]
  def dfs2(root,par):
      if par!=-1:
          dp[root]=dp[par]+1
      for i in range(1,20):
          if kthpar[root][i-1]!=-1:
              kthpar[root][i]=kthpar[kthpar[root][i-1]][i-1]
      for child in gp[root]:
          if child==par:continue
          kthpar[child][0]=root
          dfs(child,root)
          
  ans=0
  b=[]
  vis=[]
  tot=0
  time=[]
  time1=[]
  adj=[]
  mx=-1
  def dfs(a,b,p,c):
      if a==b:
          return c
      for i,j in adj[a]:
          if i==p:continue
          temp=dfs(i,b,a,c+j)
          if dfs(i,b,a):
              mx=max(i,mx)
              return 1
  eps=0.0000009
  def hnbhai(tc):
      n,k=sd()
      s=sd()
      d={}
      for i in s:
          d[i]=d.get(i,0)+1
      store={}
      for i in range(n):
          if store.get(s[i]) is None:
              store[s[i]]=[i]
          else:
              store[s[i]].append(i)
      temp=[]
      ans=[0]*n
      for i in store:
          t=store[i]
          t=[len(t)]+t
          temp.append(t)
      temp.sort(reverse=True)
      abe=[]
      tt=1
      for i in range(len(temp)):
          t=temp[i][1:]
          tot=0
          for j in t:
              abe.append([j,tt])
              if len(abe)==k:
                  for a in abe:
                      ans[a[0]]=a[1]
                  abe=[]
              tt+=1
              tot+=1
              if tt>k:
                  tt=1
              if tot==k:
                  break
      print(*ans)
  def dfs(root):
      global tot,vis,gp
      for child in gp[root]:
          if vis[child]==0:
              tot+=1
              vis[child]=1
              dfs(child)
  for _ in range(sb()):
      hnbhai(_+1)
  