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.*;
  
  public class Main {
      
      static void code(int[] arr,int n,int k) {
          Map<Integer,List<Integer>> hm=new LinkedHashMap<>();
          for(int i=0;i<n;i++) {
              if(!hm.containsKey(arr[i])) hm.put(arr[i],new ArrayList<Integer>());
              hm.get(arr[i]).add(i);
          }
          int[] res=new int[n];
          int rem=n,color=1;
          int[] freq=new int[k];
          for(int x:arr) {
              if(!hm.containsKey(x)) continue;
              List<Integer> al=hm.get(x);
              if(al.size()>=k) {
                  for(int i=1;i<=k;i++) {
                      res[al.get(i-1)]=i;
                      freq[i-1]++;
                  }
              } else {
                  for(int i=0;i<al.size();i++) {
                      if(color>k) color=1;
                      res[al.get(i)]=color;
                      freq[color-1]++;
                      color++;
                  }
              }
              hm.remove(x);
          }
          
          int min=Integer.MAX_VALUE;
          for(int i:freq) min=Math.min(min,i);
          
          
          
          for(int i=0;i<n;i++) {
              int col=res[i]-1;
              if(col>=0 && freq[col]>min) {
                  res[i]=0;
                  freq[col]--;
              }
          }
          
          for(int i:res) System.out.print(i+" ");
          System.out.println();
      }
      
      public static void main(String[] args) {
          Scanner sc=new Scanner(System.in);
          int t=sc.nextInt();
          while(t-->0) {
              int n=sc.nextInt();
              int k=sc.nextInt();
              int[] arr=new int[n];
              for(int i=0;i<n;i++) arr[i]=sc.nextInt();
              code(arr,n,k);
          }
      }
  }