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{
      public static void main(String[] args){
          Scanner sc=new Scanner(System.in);
          int t=sc.nextInt();
          while(t--!=0){
              Map<Integer,Stack<Integer>> mp=new HashMap<>();
              int n=sc.nextInt();
              int k=sc.nextInt();
              int[] tc=new int[n];
              int[] f=new int[n+5];
              int[] ans=new int[n];
              int[] kf=new int[k+1];
              int mx=n/k;
              for(int i=0;i<n;i++){
                  tc[i]=sc.nextInt();
                  f[tc[i]]++;
                  if(mp.containsKey(tc[i])==true){
                      Stack<Integer> ta=mp.get(tc[i]);
                      ta.push(i);
                      mp.put(tc[i],ta);
                  }else{
                      Stack<Integer> d=new Stack<>();
                      d.push(i);
                      mp.put(tc[i],d);
                  }
                  
              }
              int of=0;
              Stack<Integer> st=new Stack<>();
              Set<Integer> se=new HashSet<>();
              int am=0;
              for(int i=1;i<=n;i++){
                  int c=tc[i-1];
                  if(f[tc[i-1]]>=k ){
                      f[tc[i-1]]=-k;
                      kf[k]++;
                      ans[i-1]=-(f[tc[i-1]]);
                      //kf[-(f[tc[i-1]])]++;
                      f[tc[i-1]]++;
                      am++;
                  }else if( f[tc[i-1]]<0){
                      ans[i-1]=-(f[tc[i-1]]);
                      kf[-(f[tc[i-1]])]++;
                      f[tc[i-1]]++;
                      
                  }
                  else if(f[c]<k && f[c]>0){
                      of++;
                      se.add(tc[i-1]);
                      
                      //st.push(i-1);
                  }
                  
              }
              mx=Math.min(am+(of/k),mx);
              int startC=1,color=0;
              for(int ele: se){
                  int cn=0;
                  Stack<Integer> idx=mp.get(ele);
                  while(!idx.isEmpty() && cn<=k){
                      int id=idx.pop();
                      if(kf[color+1]<mx){
                          ans[id]=color+1;
                          kf[color+1]++;
                          color=(color+1)%k;
                          
                      }else cn++;
                      if(cn>k) break;
                  }
                  if(cn>k) break;
              }
              for(int i=0;i<n;i++){
                  System.out.print(ans[i]+" ");
              }
              System.out.println();
          }
      }
      
  }