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.BufferedReader;
  import java.io.InputStreamReader;
  import java.io.PrintWriter;
  import java.util.ArrayList;
  import java.util.Arrays;
  import java.util.Collections;
  import java.util.StringTokenizer;
  
  public class B2{
      public static class Pair implements Comparable<Pair> {
          int f, s;
      
          Pair(int f, int s) {
              this.f=f;
              this.s=s;
          }
      
          public int hashCode() {
              int hf = (int) (f ^ (f >>> 32));
              int hs = (int) (s ^ (s >>> 32));
              return 31 * hf + hs;
          }
      
          public boolean equals(Object o) {
              Pair other = (Pair) o;
              return s == other.s && f == other.f;
          }
      
          public int compareTo(Pair p) {
              if(this.f == p.f)
                  return Long.compare(this.s, p.s);
              return Long.compare(this.f, p.f);
          }
      
          public String toString() {
              return "(" + f + ", " + s + ")";
          }
      }
      public static void main(String[] args){
          FS sc = new FS();
          PrintWriter pw = new PrintWriter(System.out);
          int t = sc.nextInt();
          while(t-->0){
              int n = sc.nextInt();
              int k = sc.nextInt();
              int[] a = sc.nextArray(n);
              ArrayList<Pair> list = new ArrayList<>();
              for(int i=0; i<n; i++){
                  list.add(new Pair(a[i], i));
              }
              Collections.sort(list);
              int[] colors = new int[n];
              int[] filled = new int[k];
              colors[0] = 0;
              int prev = list.get(0).f;
              int count = 1, last = 1;
              filled[0]++;
              boolean unpainted = false;
              for(int i=1; i<n; i++){
                  int curr = list.get(i).f;
                  if(unpainted) colors[i] = -1;
                  if((prev==curr && count>=k)){
                      colors[i] = -1;
                      count++;
                  }
                  else if(prev==curr && count<k){
                      count++;
                      colors[i] = (last)%k;
                      filled[last%k]++;
                      last++;
                      last %= k;
                  }
                  else if(prev!=curr||count<=k){
                      colors[i] = (last)%k;
                      filled[last%k]++;
                      last++;
                      last %= k;
                      count = 1;
                  }
                  prev = curr;
              }
              int[] ans = new int[n];
              for(int i=0; i<n; i++){
                  ans[list.get(i).s] = colors[i];
              }
              int minfilled = Integer.MAX_VALUE;
              for(int fill : filled){
                  minfilled = Math.min(fill, minfilled);
              }
              for(int i=0; i<k; i++){
                  if(filled[i]>minfilled){
                      for(int j=0; j<n; j++){
                          if(filled[i]>minfilled && ans[j]==i){
                              ans[j] = -1;
                              filled[i]--;
                          }
                          else if(filled[i]<=minfilled) break;
                      }
                  }
              }
              StringBuilder sb = new StringBuilder();
              for(int i=0; i<n; i++){
                  sb.append((ans[i]+1)+" ");
              }
              System.out.println(sb);
          }
          pw.flush();
          pw.close();
      }
  
      static class FS{
          BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
          StringTokenizer st = new StringTokenizer("");
      
          String next(){
              while(!st.hasMoreElements()){
                  try{
                      st = new StringTokenizer(br.readLine());
                  } catch(Exception ignored){
                  }
              }
              return st.nextToken();
          }
          
          int[] nextArray(int n){
              int[] a = new int[n];
              for(int i = 0; i < n; i++){
                  a[i] = nextInt();
              }
              return a;
          }
          
          long[] nextLongArray(int n){
              long[] a = new long[n];
              for(int i = 0; i < n; i++){
                  a[i] = nextLong();
              }
              return a;
          }
          
          int nextInt(){
              return Integer.parseInt(next());
          }
          
          long nextLong(){
              return Long.parseLong(next());
          }
          
          double nextDouble(){
              return Double.parseDouble(next());
          }
      }
  }