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 java.util.*;
  public class Main
  {
      public static void main (String[] args)
      {
          Scanner sc=new Scanner(System.in);
          int t=sc.nextInt();
          while(t-->0)
          {
              int n=sc.nextInt();
              int k=sc.nextInt();
              int[] arr=new int[n];
              Map<Integer, List<Integer>> map=new HashMap<>();
              for(int i=0;i<n;++i)
                  arr[i]=sc.nextInt();
              for(int i=0;i<n;++i)
              {
                  if(!map.containsKey(arr[i]))
                      map.put(arr[i],new ArrayList<>());
                  map.get(arr[i]).add(i);
              }
              int count=0;
              int[] result=new int[n];
              for(int i: map.keySet())
              {
                  int temp=k;
                  List<Integer> list=map.get(i);
                  if(list.size()>=k)
                  {
                      for(int j=0;j<list.size();++j)
                      {
                          int index=list.get(j);
                          if(j<k)
                              result[index]=j+1;
                          else    
                              result[index]=0;
                      }
                  }
                  else
                  {
                      for(int j=0;j<list.size();++j)
                          count++;
                  }
              }
              int div=count-count%k;
              int p=0;
              int temp=1;
              for(int i: map.keySet())
              {
                  List<Integer> list=map.get(i);
                  if(list.size()<k)
                  {
                      if(temp>k)
                          temp=1;
                      for(int j=0;j<list.size();++j)
                      {
                          int index=list.get(j);
                          if(p<div)
                          {
                              if(temp>k)
                                  temp=1;
                              result[index]=temp;
                              ++temp;
                          }
                          else    
                              result[index]=0;
                          ++p;
                      }
                  }
              }
              for(int num: result)
                  System.out.print(num+" ");
              System.out.println();
          }
      }
  }