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 SolutionB{
         public static void main(String[] args) throws Exception{
                 Fast sc=new Fast();
          PrintWriter out=new PrintWriter(System.out);
          int t=sc.nextInt();
          while(t-->0){
        
               int n=sc.nextInt();
               int k=sc.nextInt();
  
                ArrayList<Integer>[] cnt=new ArrayList[n+1];
                for(int i=0;i<=n;i++)cnt[i]=new ArrayList<>();
  
                int[] ar=new int[n];
                for(int i=0;i<n;i++){
                  ar[i]=sc.nextInt();
                  cnt[ar[i]].add(i);
                }
                int[] kr=new int[k];
                int l=0;
                for(int i=1;i<=n;i++){
                       int p=0;
                       for(int a:cnt[i]){
                           kr[l]++;
                           l++;
                           if(l==k){
                            l=0;
                           }
                            
                           p++;
                           if(p==k)break;
                       }
  
                }
                Sort(kr);
                int max=kr[0];
                int[] ans=new int[n];
                kr=new int[k];
                l=0;
                 for(int i=1;i<=n;i++){
                      
                       int p=0;
                       for(int a:cnt[i]){
                        kr[l]++;
                         if(kr[l]<=max)ans[a]=l+1;
                            l++;
                           if(l==k){
                            l=0;
                           }
                           p++;
                           if(p==k)break;
                       }
  
                }
  
                for(int a:ans)out.print(a+" ");
                  out.println();
  
  
  
            
   
  }
       out.close();
  }
  
  static void Sort(int[] ar){
      ArrayList<Integer> al=new ArrayList<>();
      for(int i=0;i<ar.length;i++){al.add(ar[i]);}
      Collections.sort(al);
      for(int i=0;i<al.size();i++)ar[i]=al.get(i);
    }
  static int gcd(int a,int b){
      if(b==0) return a;
      else return  gcd(b,a%b);
  }
  }
  class Pair{
      int x, y;
       Pair(int x,int y){
          this.x=x;
          this.y=y;
       }
  }
  class Fast{
    BufferedReader br; 
    StringTokenizer st;
    public Fast(){ br=new BufferedReader(new InputStreamReader(System.in)); }
    String next() 
          {  while (st == null || !st.hasMoreElements()) 
              {   try
                  {     st = new StringTokenizer(br.readLine()); } 
                  catch (IOException  e) 
                  {    e.printStackTrace(); } 
              } 
              return st.nextToken(); 
          }
    int nextInt(){  return Integer.parseInt(next()); }
    long nextLong(){     return Long.parseLong(next()); } 
    double nextDouble() {     return Double.parseDouble(next()); } 
    String nextLine() 
          { 
              String str = ""; 
              try{      str = br.readLine(); } 
              catch (IOException e)  {  e.printStackTrace(); } 
              return str; 
          } 
  }
      