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.io.*;
  
  public class B2_Wonderful_Coloring_2{
      public static void main(String[] args) {
          FastScanner s= new FastScanner();
          StringBuilder res = new StringBuilder();
          int t=s.nextInt();
          int p=0;
          while(p<t){
          int n=s.nextInt();
          int k=s.nextInt();
          int array[]= new int[n];
          for(int i=0;i<n;i++){
              array[i]=s.nextInt();
          }
          HashMap<Integer,ArrayList<Integer>> map = new HashMap<Integer,ArrayList<Integer>>();
          ArrayList<Integer> unique = new ArrayList<Integer>();
          for(int i=0;i<n;i++){
              if(map.containsKey(array[i])){
                  map.get(array[i]).add(i);
              }
              else{
                  ArrayList<Integer> obj = new  ArrayList<Integer>();
                  obj.add(i);
                  map.put(array[i],obj);
                  unique.add(array[i]);
              }
          }
          int ans[]= new int[n];
          int left=n;
          HashMap<Integer,Integer> done = new HashMap<Integer,Integer>();
          for(int i=0;i<unique.size();i++){
                int num=unique.get(i);
                if(map.get(num).size()>=k){
                    done.put(num,1);
                    for(int j=0;j<k;j++){
                        int hh=map.get(num).get(j);
                        ans[hh]=j+1;
                    }
                    left-=map.get(num).size();
  
                }
          }
          // int val=1;
          // for(int i=0;i<unique.size();i++){
          //     if(left<k){
          //         break;
          //     }
          //     int num=unique.get(i);
          //     if(done.containsKey(num)){
          //         continue;
          //     }
          //     for(int j=0;j<map.get(num).size();j++){
          //         int hh=map.get(num).get(j);
          //         ans[hh]=val;
          //         val++;
          //         if(val==(k+1)){
          //             val=1;
          //         }
          //         left--;
                  
  
          //     }
  
  
          // }
          // for(int i=0;i<n;i++){
          //     System.out.print(ans[i]+" ");
          // }
          // System.out.println(left);
         // int j=0;
         int i=0;
         int start=0;
          while(left>=k){
               int value=1;
              // for(;j<n;j++){
              //     int number=array[j];
              //     if(done.containsKey(number)){
              //     //    System.out.print(number+" ");
              //         continue;
              //     }
              //   //  System.out.println("vali"+value+"j "+j);
              //     ans[j]=value;
              //     value++;
              //     if(value==(k+1)){
              //         j++;
              //         break;
              //     }
              // }
              // left-=k;
              for( ;i<unique.size();i++){
                  int number=unique.get(i);
                 if(done.containsKey(number)){
                     continue;
                 }
                 int flag=0;
                 for(int j=0;j<map.get(number).size();j++){
                     int yo =map.get(number).get(j);
                     if(ans[yo]!=0){
                         continue;
                     }
                     ans[yo]=value;
                     value++;
                     if(value==k+1){
                         flag=1;
                         break;
                     }
                 }
                 if(flag==1){
                     break;
                 }
              }
              left-=k;
  
          }
          for(int l=0;l<n;l++){
              res.append(ans[l]+" ");
          }
          res.append("\n");
          p++;
      }
      System.out.println();
      System.out.println(res);
  
  
      }
      static class FastScanner {
  		BufferedReader br;
  		StringTokenizer st;
   
  		public FastScanner(String s) {
  			try {
  				br = new BufferedReader(new FileReader(s));
  			} catch (FileNotFoundException e) {
  				// TODO Auto-generated catch block
  				e.printStackTrace();
  			}
  		}
   
  		public FastScanner() {
  			br = new BufferedReader(new InputStreamReader(System.in));
  		}
   
  		String nextToken() {
  			while (st == null || !st.hasMoreElements()) {
  				try {
  					st = new StringTokenizer(br.readLine());
  				} catch (IOException e) {
  					// TODO Auto-generated catch block
  					e.printStackTrace();
  				}
  			}
  			return st.nextToken();
  		}
   
  		int nextInt() {
  			return Integer.parseInt(nextToken());
  		}
   
  		long nextLong() {
  			return Long.parseLong(nextToken());
  		}
   
  		double nextDouble() {
  			return Double.parseDouble(nextToken());
  		}
  	}
  
  
  
  
  }