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.*;
  
  /* Name of the class has to be "Main" only if the class is public. */
  
  public class Codechef
  {
  	public static void main (String[] args) throws java.lang.Exception
  	{
  	    try{
  	    FastReader read=new FastReader();
  		// your code goes here
  		StringBuffer sb = new StringBuffer();
  		int t=read.nextInt();
  		while(t>0)
  		{
  		   
  		   int n = read.nextInt();
  		   int k = read.nextInt();
  		   int c[]=new int[k];
  		   int cc[]=new int[k];
  		   pair []p=new pair[n];
  		   for(int i=0;i<n;i++)
  		   {
  		       int y=read.nextInt();
  		       p[i]=new pair(y,i);
  		   }
  		   int sett=1;
  		   int a[]=new int[200003];
  		    Arrays.sort(p);
  		    int ans[]=new int[n];
  		    for(int i=0;i<n;i++)
  		    {
  		        a[p[i].val]++;
  		        if(a[p[i].val]>k)
  		        {
  		            p[i].set=0;
  		        }
  		        else
  		        {
  		            p[i].set=sett;
  		            ans[p[i].ind]=p[i].set;
  		            c[sett-1]++;
  		            cc[sett-1]++;
  		            sett++;
  		            if(sett>k)
  		            sett=1;
  		        }
  		    }
  		    Arrays.sort(c);
  		    int min = c[0];
  		    //System.out.println("y");
  		    for(int i=0;i<n;i++)
  		    {
  		        if(ans[i]!=0)
  		        {
  		        if(cc[ans[i]-1]<=min)
  		        sb.append(ans[i]+" ");
  		        else
  		        {
  		            sb.append("0"+" ");
  		            cc[ans[i]-1]--;
  		        }
  		        }
  		        else
  		        {
  		             sb.append("0"+" ");
  		        }
  		    }
  		   
  		    sb.append('\n');
  		    t--;
  		}
  		System.out.println(sb);
  	    }
  	    catch(Exception e)
  	    {return;
  	    }
  	}
  	
  	
  	 static class FastReader
      {
          BufferedReader br;
          StringTokenizer st;
   
          public FastReader()
          {
              br = new BufferedReader(new
                       InputStreamReader(System.in));
          }
   
          String next()
          {
              while (st == null || !st.hasMoreElements())
              {
                  try
                  {
                      st = new StringTokenizer(br.readLine());
                  }
                  catch (IOException  e)
                  {
                      e.printStackTrace();
                  }
              }
              return st.nextToken();
          }
   
          int nextInt()
          {
              return Integer.parseInt(next());
          }
   
          long nextLong()
          {
              return Long.parseLong(next());
          }
   
          double nextDouble()
          {
              return Double.parseDouble(next());
          }
   
          String nextLine()
          {
              String str = "";
              try
              {
                  str = br.readLine();
              }
              catch (IOException e)
              {
                  e.printStackTrace();
              }
              return str;
          }
      }
  }
  class pair implements Comparable<pair>
  {
      int val;
      int ind;
      int set;
      pair(int v,int i)
      {
          val=v;
          ind=i;
          set=0;
      }
      public int compareTo(pair p)
      {
          return this.val-p.val;
      }
  }
  