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: // Don't place your source in a package
  
  
  import javax.swing.*;
  import java.lang.reflect.Array;
  import java.text.DecimalFormat;
  import java.util.*;
  import java.lang.*;
  import java.io.*;
  import java.math.*;
  import java.util.stream.Stream;
  
  
  
  // Please name your class Main
  public class Main {
      static FastScanner fs=new FastScanner();
      static class FastScanner {
          BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
          StringTokenizer st=new StringTokenizer("");
          public String next() {
              while (!st.hasMoreElements())
                  try {
                      st=new StringTokenizer(br.readLine());
                  } catch (IOException e) {
                      e.printStackTrace();
                  }
              return st.nextToken();
          }
          int Int() {
              return Integer.parseInt(next());
          }
  
          long Long() {
              return Long.parseLong(next());
          }
  
          String Str(){
              return next();
          }
      }
  
  
      public static void main (String[] args) throws java.lang.Exception {
          PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
          //reading /writing file
          //Scanner sc=new Scanner(new File("input.txt"));
          //PrintWriter pr=new PrintWriter("output.txt");
  
  
          int T=Int();
          for(int t=0;t<T;t++){
              int n=Int();
              int k=Int();
              int A[]=Arr(n);
              Solution sol=new Solution(out);
              sol.solution(A,k);
          }
          out.close();
  
      }
  
  
  
  
      public static int[] Arr(int n){
          int A[]=new int[n];
          for(int i=0;i<n;i++)A[i]=Int();
          return A;
      }
      public static int Int(){
          return fs.Int();
      }
      public static long Long(){
          return fs.Long();
      }
      public static String Str(){
          return fs.Str();
      }
  
  }
  
  
  
  
  
  
  
  
  class Solution {
      PrintWriter out;
      int INF = Integer.MAX_VALUE;
      int MOD = 1000000007;
      int mod = MOD;
      public Solution(PrintWriter out) {
          this.out = out;
      }
  
  
  
  
      public void solution(int A[],int k) {
          int res[]=new int[A.length];
          int B[][]=new int[A.length][2];
          int cnt[]=new int[A.length+10];
          int sum=0;
          for(int i=0;i<A.length;i++){
              B[i][0]=A[i];
              B[i][1]=i;
              cnt[A[i]]++;
          }
          for(int i:cnt){
              sum+=Math.min(i,k);
          }
  
  
  
          Arrays.sort(B,(a,b)->{
              return a[0]-b[0];
          });
  
          long l =1,r=A.length;
          long use=-1;
          while(l<=r){
              long mid=l+(r-l)/2;
              if(sum>=mid*k){
                  use=mid;
                  l=mid+1;
              }
              else{
                  r=mid-1;
              }
          }
  
          Queue<Integer>q=new LinkedList<>();
          for(int i=0;i<use;i++){
              for(int j=1;j<=k;j++){
                  q.add(j);
              }
          }
  
          // System.out.println(use);
          //  System.out.println(Arrays.toString(cnt));
  
          for(int i=0;i<B.length;i++){
              int j = i;
              int x =0;
              while(j<B.length&&B[j][0]==B[i][0]){
                  j++;
                  x++;
              }
  
              x=Math.min(x,k);
              for(int y=0;y<x;y++){
                  if(q.size()==0)break;
                  int index=B[i+y][1];
                  res[index]=q.poll();
              }
  
              i=j-1;
          }
  
  
          for(int i:res){
              out.print(i+" ");
          }out.println();
      }
  
  
  
  }
  