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 Solution {
  
      public static long gcd(long a, long b){
          if(b==0){
              return a;
          }
          return gcd(b, a%b);
      }
  
      public static long gcdSum(long b){
          long a = 0;
  
          long temp = b;
          while(temp!=0){
              a = a + temp%10;
              temp = temp/10;
          }
          return gcd(a,b);
  
      }
  
  
      public  static class inputPair implements Comparable<inputPair>{
          private Integer a;
          private Integer b;
  
          public inputPair(int temp, int j) {
              a = temp;
              b = j;
          }
          public int compareTo(inputPair o1) {
              return this.a-o1.a;
          }
  
      }
  
  
  
  
      public static void main(String[] args){
          Scanner s = new Scanner(System.in);
          int t = s.nextInt();
          //long  factArray[] = new long [42];
  
  
          for(int i =0;i<t;i++) {
              int n = s.nextInt();
              int k = s.nextInt();
  
              inputPair arr[] = new inputPair[n];
              int orig[] = new int[n];
              Set<inputPair> m[] = new HashSet[k+1];
              Set<Integer> dupm[] = new HashSet[k+1];
  
              for(int j =0;j<=k;j++){
                  m[j] = new HashSet<>();
                  dupm[j] = new HashSet<>();
              }
  
              for(int j =0;j<n;j++){
                  int p = s.nextInt();
                  arr[j] = new inputPair(p,j);
                  //orig[j] = s.nextInt();
              }
              Arrays.sort(arr);
  //            for(int j =0;j<n;j++){
  //                System.out.println(arr[j].a + "==" + arr[j].b);
  //            }
  
              int temp  = 0;
              //m[temp].add(arr[0]);
              //temp = (temp+1)%k;
  
              for(int j =0;j<n;j++){
                  if(!dupm[temp].contains(arr[j].a)) {
                      m[temp].add(arr[j]);
                      dupm[temp].add(arr[j].a);
                      //System.out.println(arr[j].a+"->"+temp);
                      temp = (temp + 1) % k;
                  }
                  else {
                      m[k].add(arr[j]);
                      dupm[k].add(arr[j].a);
                      //System.out.println(arr[j].a+"-->"+k);
                  }
  
              }
  
              int min=10000000;
              for(int j =0;j<k+1;j++){
                  if(j<k&&m[j].size()<min)
                      min = m[j].size();
              }
  
              for(int j =0;j<k+1;j++){
                  int sizeCurrent = 0;
                  for(inputPair current : m[j]){
                      if(sizeCurrent>=min){
                          orig[current.b] = 0;
                          continue;
                      }
                      orig[current.b] = j+1;
                      sizeCurrent++;
                  }
              }
  
              for(int j =0;j<n;j++){
                  if(orig[j]==k+1)
                      orig[j] = 0;
                  System.out.print(orig[j]+" ");
              }
              System.out.println();
          }
  
  
          return;
      }
  }
  