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 javafx.util.Pair;
  
  import java.io.*;
  import java.util.*;
  public class Main {
      public static void main(String args[])
      {
          FastReader input=new FastReader();
          PrintWriter out=new PrintWriter(System.out);
          int T=input.nextInt();
          while(T-->0)
          {
              int n=input.nextInt();
              int k=input.nextInt();
              int a[]=new int[n];
              ArrayList<Pair> list1=new ArrayList<>();
              for(int i=0;i<n;i++)
              {
                  a[i]=input.nextInt();
                  list1.add(new Pair(a[i],i));
              }
              Collections.sort(list1, new Comparator<Pair>() {
                  @Override
                  public int compare(Pair o1, Pair o2) {
                      return (int)o1.getKey()-(int)o2.getKey();
                  }
              });
              int color[]=new int[n];
              int x=1;
              int count=1;
              ArrayList<Integer> freq=new ArrayList<>();
              for(int i=0;i<list1.size()-1;i++)
              {
                  if((int)list1.get(i).getKey()!=(int)list1.get(i+1).getKey())
                  {
                      freq.add(count);
                      count=1;
                  }
                  else
                  {
                      count++;
                  }
              }
              freq.add(count);
              int y=0;
              int z=0;
              int k1=k;
              int max=n-n%k1;
              int sum=0;
              for(int i=0;i<freq.size();i++)
              {
                  sum+=Math.min(k1,freq.get(i));
              }
              max=sum-sum%k1;
              for(int i=0;i<freq.size();i++)
              {
                  int v=freq.get(i);
                  for(int j=0;j<v;j++)
                  {
                      if(j<k1 && z<max)
                      {
                          int in=(int)list1.get(y).getValue();
                          color[in]=x;
                          x++;
                          if(x>k1) x=1;
                          z++;
                      }
                      y++;
                  }
              }
              for(int i=0;i<n;i++)
              {
                  out.print(color[i]+" ");
              }
              out.println();
          }
          out.close();
      }
      static class FastReader
      {
          BufferedReader br;
          StringTokenizer st;
          public FastReader()
          {
              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;
          }
      }
  }