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 Main
  {
  	public static void main(String ar[]) {
  			Scanner sc = new Scanner(System.in);
  			int q = sc.nextInt();
  			while(q--!=0) {
  				int n = sc.nextInt();
  				int k = sc.nextInt();
  				int paintedarr[] = new int[n];
  				int arr[] = new int[n];
  				int freqcol[] = new int[k+1];
  				HashMap<Integer,Integer> hm = new HashMap<>();
  				HashMap<Integer,List<Integer>> hmm = new HashMap<>();
  				for(int i=0;i<n;i++) {
  					arr[i] = sc.nextInt();
  					if(hm.containsKey(arr[i])) {
  						List<Integer> temp = hmm.get(arr[i]);
  						temp.add(i);
  						hm.put(arr[i],hm.get(arr[i])+1);
  						hmm.put(arr[i],temp);
  					}else {
  						hm.put(arr[i],1);
  						hmm.put(arr[i],new ArrayList<>());
  						List<Integer> temp = hmm.get(arr[i]);
  						temp.add(i);
  						hmm.put(arr[i],temp);
  					}
  				}
  				int colorToPaint = 1;
  				for(int i=0;i<n;i++) {
  					if(hm.containsKey(arr[i])) {
  						int freqOfEle = hm.get(arr[i]);
  						if(freqOfEle>=k) {
  							List<Integer> tmp = hmm.get(arr[i]);
  							int zz = 0;
  							for(int v=1;v<=k;v++) {
  							   
  								freqcol[v]++;
  								paintedarr[tmp.get(zz++)] = v;
  							}
  							hmm.remove(arr[i]);
  							hm.remove(arr[i]);
  						}else {
  							List<Integer> tmp = hmm.get(arr[i]);
  							int ssize = tmp.size();
  							for(int v=0;v<ssize;v++) {
  								if(colorToPaint==k+1) {
  									colorToPaint = 1;
  									
  									freqcol[colorToPaint]++;
  									paintedarr[tmp.get(v)] = colorToPaint++;
  								}else {
  								    
  								    freqcol[colorToPaint]++;
  									paintedarr[tmp.get(v)] = colorToPaint++;
  								}
  							}
  							hmm.remove(arr[i]);
  							hm.remove(arr[i]);
  						}
  					}else {
  						continue;
  					}
  				}
  				int findMinCol = Integer.MAX_VALUE;
  				for(int i=1;i<=k;i++) {
  				    if(freqcol[i]<findMinCol) {
  				        findMinCol = freqcol[i];
  				    }
  				}
  				for(int i=0;i<n;i++) {
  				    if(freqcol[paintedarr[i]]>findMinCol) {
  				        System.out.print(0+" ");
  				        freqcol[paintedarr[i]]--;
  				    }else {
  				        System.out.print(paintedarr[i]+" ");
  				    }
  				     
  					
  				}
  				
  				System.out.println();
  
  				
  
  			}
  		}
  }
  