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: /*
   * To change this license header, choose License Headers in Project Properties.
   * To change this template file, choose Tools | Templates
   * and open the template in the editor.
   */
  
  import java.util.*;
    import java.io.*;
    public class A
      {
          static BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
          static BufferedWriter bw=new BufferedWriter(new OutputStreamWriter(System.out));
          static StringTokenizer st=null;
          static String next()
          {
              while(st==null || !st.hasMoreElements())
              {
                  try{
                      st=new StringTokenizer(br.readLine());
                  } catch (IOException e){
                      e.printStackTrace();
                  }
              }
              return st.nextToken();
          }
          static int nextInt()
          {
              return Integer.parseInt(next());
          }
          static long nextLong()
          {
              return Long.parseLong(next());
          }
          
          static double nextDouble(){
              return Double.parseDouble(next());
          }
          
       
         public static void main(String[] args) {
              try{
                  
                  int t=nextInt();
                  for(int i=0;i<t;i++){
                      
                      int n=nextInt();
                      int k=nextInt();
                      
                      int a[]=new int[n];
                      int nums[]=new int[n+1];
                      HashMap<Integer,TreeSet<Integer>> hmpos=new HashMap<>();
                      
                      for(int j=0;j<n;j++){
                          a[j]=nextInt();
                          if(hmpos.get(a[j])==null){
                              TreeSet<Integer> ts=new TreeSet<>();
                              ts.add(j);
                              hmpos.put(a[j], ts);
                          }else{
                              TreeSet<Integer> ts=hmpos.get(a[j]);
                              ts.add(j);
                              hmpos.put(a[j], ts);
                          }
                          nums[a[j]]++;
                      }
                      
                      int ans[]=new int[n];
                      TreeSet<Integer> tsk=new TreeSet<>();
                      
                      for(int j=0;j<n;j++){
                          
                          if(nums[a[j]]>=k){
                              nums[a[j]]=k;
                              tsk.add(a[j]);
                          }
                          
                          if(nums[a[j]]>0 && tsk.contains(a[j])==true){
                              int set=nums[a[j]];
                              ans[j]=set;
                              nums[a[j]]--;
                          }
                      }
                      
                      int remaining=0;
                      TreeSet<Integer> remainingNum=new TreeSet<>();
                      
                      for(int j=0;j<n;j++){
                          if(nums[a[j]]>0 && remainingNum.contains(a[j])==false){
                              
                             remainingNum.add(a[j]);
                             remaining+=nums[a[j]];
                          }
                      }
                      
                      int eachSet=remaining/k;
                      int set=1;
                      int count=1;
                      int totalnum=0;
                      
                      while(eachSet>0 && !remainingNum.isEmpty()){
                          int num=remainingNum.pollFirst();
                          
                          
                          TreeSet<Integer> posts=hmpos.get(num);
                          while(!posts.isEmpty()){
                              int pos=posts.pollFirst();
                              ans[pos]=set;
                              count++;
                              totalnum++;
                              set++;
                              if(totalnum==(eachSet*k)){
                                  break;
                              }
                              
                              if(set>k){
                                  set=1;
                              }
                             /* if(count>eachSet){
                                  count=1;
                              }*/
                          }
                          hmpos.put(num, posts);
                          
                          if(totalnum==(eachSet*k)){
                              break;
                          }
                          
                      }
                      
                      for(int j=0;j<n;j++){
                          bw.write(ans[j]+" ");
                      }
                      bw.write("\n");
                     
                      
                      
                  }
                  
      		bw.flush();
              } catch (Exception e) {
                  e.printStackTrace();
              }
          }
      }  