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.lang.*;
  import java.io.*;
  import java.math.*;
  
  /* Name of the class has to be "Main" only if the class is public. */
  public final class codechef {
  	public static boolean found=false;
  	public static void main(String[] args) throws java.lang.Exception {
  		// your code goes here
  		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  		int t = Integer.parseInt(br.readLine());
  		while (t-- > 0) {
  			String a[]=br.readLine().split(" ");
  			int n=Integer.parseInt(a[0]);
  			int k=Integer.parseInt(a[1]);
  			String b[]=br.readLine().split(" ");
  			int arr[]=new int[n];
  			for(int i=0;i<n;i++)arr[i]=Integer.parseInt(b[i]);
  			int cnf=0;
  			int paint[]=new int[n];
  			HashMap<Integer,List<Integer>>res=new HashMap<>();
  			for(int i=0;i<n;i++){
  				if(res.containsKey(arr[i])){
  				    List<Integer> tem=res.get(arr[i]);
  				    tem.add(i);
  					res.put(arr[i],tem);
  				}
  				else{
  					ArrayList<Integer> temp=new ArrayList<>();
  					temp.add(i);
  					 res.put(arr[i],temp);
  				}
  			}
  			int curr=1;
  			int count=0;
  			Queue<Integer> q=new LinkedList<>();
  			for (Map.Entry<Integer,List<Integer>> gg : res.entrySet()){
  				List<Integer> ind=gg.getValue();
  				if(ind.size()>=k){
  					for(int i=0;i<k;i++)paint[ind.get(i)]=i+1;
  				
  				}
  				else{
  				    for(int f:ind)q.add(f);
  				}
  				
  			}
  			if(q.size()>=k){
  			    int counter=q.size()/k;
  			    for(int i=0;i<counter*k;i++){
  			        paint[q.poll()]=curr;
  			        curr++;
  			        if(curr>k)curr=1;
  			    }
  			}
  			for(int i=0;i<n;i++)System.out.print(paint[i]+" ");
  			System.out.println();
  			
  		}		
  		//System.out.flush();
  	}
  }