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.*;
  
  import java.lang.Math;
  
  public class B2_Wonderful_Coloring_2{
  
  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;
      }
  
      int[] readintarray(int n) {
          int res [] = new int[n];
          for(int i = 0; i<n; i++)
          res[i] = nextInt();
          return res;
      }
  
      long[] readlongarray(int n) {
          long res [] = new long[n];
          for(int i = 0; i<n; i++)
          res[i] = nextLong();
          return res;
      }
  
  }
  
      static int gcd(int a, int b) {
          if (a == 0)
              return b;
          return gcd(b % a, a);
      }
  
      static int lcm(int a, int b) {
           return (a / gcd(a, b)) * b;
      }
  
      public static void main(String args[]) throws IOException {
          FastReader s=new FastReader();
          int t;
          t =s.nextInt();
          while (t-- > 0) {
  
              int n=s.nextInt();
              int k=s.nextInt();
              int[] a=s.readintarray(n);
             
              HashMap<Integer,Integer> freq=new HashMap<Integer,Integer>();
              HashMap<Integer,Vector<Integer>> indexmap=new HashMap<Integer,Vector<Integer>>();
  
              for(int i=0;i<n;i++){
                  if(freq.containsKey(a[i])){
                    if(freq.get(a[i])<k){
                          freq.put(a[i],freq.get(a[i])+1);
                          Vector<Integer> indices=indexmap.get(a[i]);
                          indices.add(i);
                          indexmap.put(a[i],indices);
  
                      }  
                  }
                  else{
                      freq.put(a[i],1);
                      Vector<Integer> indices=new Vector<Integer>();
                      indices.add(i);
                      indexmap.put(a[i],indices);
  
                  }
              }
              // ans[10] = {0};
              // color = 1;
  
              // 3 -> 0 6
              //      1 2
              // 1 -> 1 2 3
              //      3 1 2    
              // 10-> 5 7 8 
              //      3 1 2
              // 2 -> 9
              //      3
  
                   
  
              //      ans =  [1,3,1,2,0,3,2,1,2,3]
              int color=1;
              int[] ans=new int[n];
              Arrays.fill(ans,0);
  
              for(Map.Entry<Integer,Vector<Integer>> m: indexmap.entrySet()){
                  Vector<Integer> index=(Vector<Integer>)m.getValue();
                  for(int i=0;i<index.size();i++){
                      ans[index.get(i)]=color;
                      color++;
                      if(color>k)
                      color=1;
                  }
  
              }
              
              // System.out.println(color);
              for(int i=1;i<=color-1;i++){
                  for(int j=0;j<n;j++){
                      if(ans[j]==i){
                          ans[j]=0;
                          break;
                      }
                  }
              }
  
              for(int i=0;i<n;i++){
                  System.out.print(ans[i]+" ");
              }
              System.out.println();
          }
      }
  }