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: /* package codechef; // don't place package name! */
  
  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
  {
      static class Element{
          int color, value, index;
          Element(int v, int i){
              this.color = 0;
              this.value = v;
              this.index = i;
          }
      }
  	public static void main (String[] args) throws java.lang.Exception
  	{
  	   
  	    Scanner sc = new Scanner(System.in);
  		int t = sc.nextInt();
  		while(t-->0)
  		{
  		   int n = sc.nextInt();
  		   int k = sc.nextInt();
  		   int count[] = new int[n+1];
  		   Element arr[] = new Element[n];
  		   for(int i = 0; i<n; i++)
  		   {
  		       arr[i] = new Element(sc.nextInt(), i);
  		   }
  		   Arrays.sort(arr, new Comparator<Element>(){
  		       public int compare(Element a, Element b)
  		       {
  		           return a.value-b.value;
  		       }
  		   });
  		   
  		   int j = 1;
  		   for(int i = 0; i<n; i++)
  		   {
  		       if(count[arr[i].value]<k)
  		       {
  		           arr[i].color = j;
  		           j = (j+1)%(k+1);
  		           if(j==0) j = 1;
  		           count[arr[i].value]++;
  		       }
  		   }
  		   if(j!=1)
  		   {
  		       int i = n-1;
  		       while(i>=0&&arr[i].color!=k)
  		       {
  		           arr[i].color = 0;
  		           i--;
  		       }
  		   }
  		   Arrays.sort(arr, new Comparator<Element>(){
  		       public int compare(Element a, Element b)
  		       {
  		           return a.index-b.index;
  		       }
  		  });
  		  for(int i = 0; i<n; i++)
  		  {
  		      System.out.print(arr[i].color+" ");
  		  }
  		  System.out.println();
  		}
  		
  	}
  }
  