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.*;
  import java.util.stream.Collectors;
  import java.lang.*;
  import java.io.*;
   
  public class Problem {
  	public static int[] color(int n,int k,int[] ar)
  	{
  		int[] ans = new int[n];
  		Map<Integer,List<Integer>> freq = new HashMap<Integer,List<Integer>>();
  		for(int i=0;i<n;i++)
  		{
  			if(freq.containsKey(ar[i]))
  				freq.get(ar[i]).add(i);
  			else
  			{
  				List<Integer> indices = new ArrayList<Integer>();indices.add(i);
  				freq.put(ar[i], indices);
  			}
  		}
  		List<Integer> tempkeys = new ArrayList<Integer>();
  		int tempcount=0;
  		for(Map.Entry<Integer, List<Integer>> entry : freq.entrySet())
  		{
  			if(entry.getValue().size()>=k)
  			{
  				List<Integer> indices = entry.getValue();
  				int col=1;
  				for(int i=0;i<k;i++,col++)
  					ans[indices.get(i)]=col;
  			}
  			else
  			{
  				tempkeys.add(entry.getKey());
  				tempcount+=entry.getValue().size();
  				if(tempcount>=k)
  				{
  					int col=1;
  					for(int i=0;i<tempkeys.size()-1 && col<=k;i++)
  					{
  						List<Integer> indices = freq.get(tempkeys.get(i));
  						for(int j=0;j<indices.size() && col<=k;j++,col++)
  							ans[indices.get(j)]=col;
  					}
  					List<Integer> indices = freq.get(tempkeys.get(tempkeys.size()-1));
  					int p=0;
  					while(col<=k && p<indices.size())
  					{
  						ans[indices.get(p)]=col;
  						col++;
  						p++;
  					}
  					int lastkey = tempkeys.get(tempkeys.size()-1);
  					tempcount=0;
  					tempkeys.clear();
  					if(p<indices.size())
  					{
  						tempkeys.add(lastkey);
  						indices = indices.subList(p,indices.size());
  						freq.replace(lastkey, indices);
  						tempcount += indices.size();
  					}
  				}
  			}
  		}
  		
  		return ans;
  	}
  	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[] num = new int[n];
  			for(int i=0;i<n;i++)
  				num[i]=sc.nextInt();
  			int[] ans = color(n,k,num);
  			for(int i=0;i<n;i++)
  				System.out.print(ans[i]+" ");
  			System.out.println();
  		}
  	}
  }