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.io.*;
  import java.util.*;
  
  public class CF_1551b2{
      public static final void main(String[] args){
          Kattio io= new Kattio();
          int t= io.getInt();
          while(t-->0){
              int n= io.getInt(), k= io.getInt();
              int[] a= new int[n];
              int[] f= new int[n+1];
              int[][] use= new int[n+1][];
              int[] idx= new int[n+1];
              for(int i=0; i<n; i++){
                  a[i]= io.getInt();
                  f[a[i]]++;
              }
              //err("f=%s", Arrays.toString(f));
              int nonFull= 0;                     // need to have same amount of all colors
              for(int v=1; v<=n; v++){
                  if(f[v]<k) nonFull+= f[v];      // for those with f[v]>=k we'll always use k colors
              }
              int nonFullColored= k*(nonFull/k);  // for those with f[v]>l we'll allocate colors in loop
              
              //err("n=%d, k=%d, nonFull=%d, nonFullColored=%d", n, k, nonFull, nonFullColored);
  
              // allocate colors to values
              for(int v=1, used=0, c=1; v<=n; v++){
                  if(f[v]>=k){ // -- first k colored
                      use[v]= new int[k+1];
                      for(int i=1; i<=k; i++){
                          use[v][++idx[v]]= i;
                      }
                  }else if(f[v]>0 && used<nonFullColored){ //f[v]<k -- nonFullColored
                      use[v]= new int[f[v]+1];
                      for(int i=1; i<=f[v] && used<nonFullColored; i++){
                          use[v][++idx[v]]= c++;
                          if(c>k) c=1;
                          used++;
                      }
                  }else if(f[v]>0){
                      use[v]= new int[]{0};
                  }
              }   
              // create colored array
              StringBuilder ans= new StringBuilder();
              for(int i=0; i<n; i++){
                  int v= a[i];
                  int c= use[v][idx[v]];
                  if(idx[v]>0) --idx[v];
                  ans.append(c);
                  if(i!=n-1) ans.append(' ');
              }                                 
              io.println(ans);
          }
          io.close();
      }
  
      static class Kattio extends PrintWriter {
          private BufferedReader r;
          private String line, token;
          private StringTokenizer st;
      
          public Kattio(){this(System.in);}
          public Kattio(InputStream i){
          super(new BufferedOutputStream(System.out));
              r= new BufferedReader(new InputStreamReader(i));
          }
          public Kattio(InputStream i, OutputStream o){
          super(new BufferedOutputStream(o));
              r= new BufferedReader(new InputStreamReader(i));
          }
          public boolean isLineEnd(){
              return !st.hasMoreTokens();
          }
          public boolean hasMoreTokens(){
              return peekToken()!=null;
          }
          public int getInt(){
              return Integer.parseInt(nextToken());
          }
          public double getDouble(){ 
              return Double.parseDouble(nextToken());
          }
          public long getLong(){
              return Long.parseLong(nextToken());
          }
          public String getWord(){
              return nextToken();
          }
          private String peekToken(){
              if(token==null) try {
                  while(st==null || !st.hasMoreTokens()) {
                      line= r.readLine();
                      if(line==null) return null;
                      st= new StringTokenizer(line);
                  }
                  token= st.nextToken();
              }catch(IOException e){}
              return token;
          }
          private String nextToken() {
              String ans= peekToken();
              token= null;
              return ans;
          }
      }    
  }