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.*;
  import java.util.*;
  
  public class Main{
  	static void main() throws Exception{
  		int n=sc.nextInt(),k=sc.nextInt();
  		LinkedList<Integer>[]occ=new LinkedList[n];
  		for(int i=0;i<n;i++)occ[i]=new LinkedList<Integer>();
  		
  		for(int i=0;i<n;i++)occ[sc.nextInt()-1].add(i);
  		
  		int[]ans=new int[n];
  		LinkedList<Integer>cur=new LinkedList<Integer>();
  		for(int i=0;i<n;i++) {
  			for(int j=0;j<k && !occ[i].isEmpty();j++) {
  				cur.add(occ[i].pollFirst());
  				if(cur.size()==k) {
  					for(int o=0;o<k;o++) {
  						ans[cur.poll()]=o+1;
  					}
  				}
  			}
  		}
  		
  		for(int i=0;i<n;i++)pw.print(ans[i]+" ");
  		pw.println();
  	}
  	public static void main(String[] args) throws Exception{
  		sc=new MScanner(System.in);
  		pw = new PrintWriter(System.out);
  		int tc=1;
          tc=sc.nextInt();
  		for(int i=1;i<=tc;i++) {
  //			pw.printf("Case #%d: ", i);
  			main();
  		}
  		
  		pw.flush();
  	}
  	static PrintWriter pw;
  	static MScanner sc;
  	static class MScanner {
  		StringTokenizer st;
  		BufferedReader br;
  		public MScanner(InputStream system) {
  			br = new BufferedReader(new InputStreamReader(system));
  		}
  	 
  		public MScanner(String file) throws Exception {
  			br = new BufferedReader(new FileReader(file));
  		}
  	 
  		public String next() throws IOException {
  			while (st == null || !st.hasMoreTokens())
  				st = new StringTokenizer(br.readLine());
  			return st.nextToken();
  		}
  		public int[] intArr(int n) throws IOException {
  	        int[]in=new int[n];for(int i=0;i<n;i++)in[i]=nextInt();
  	        return in;
  		}
  		public long[] longArr(int n) throws IOException {
  	        long[]in=new long[n];for(int i=0;i<n;i++)in[i]=nextLong();
  	        return in;
  		}
  		public int[] intSortedArr(int n) throws IOException {
  	        int[]in=new int[n];for(int i=0;i<n;i++)in[i]=nextInt();
  	        shuffle(in);
  	        Arrays.sort(in);
  	        return in;
  		}
  		public long[] longSortedArr(int n) throws IOException {
  	        long[]in=new long[n];for(int i=0;i<n;i++)in[i]=nextLong();
  	        shuffle(in);
  	        Arrays.sort(in);
  	        return in;
  		}
  		public Integer[] IntegerArr(int n) throws IOException {
  	        Integer[]in=new Integer[n];for(int i=0;i<n;i++)in[i]=nextInt();
  	        return in;
  		}
  		public Long[] LongArr(int n) throws IOException {
  	        Long[]in=new Long[n];for(int i=0;i<n;i++)in[i]=nextLong();
  	        return in;
  		}
  		public String nextLine() throws IOException {
  			return br.readLine();
  		}
  	 
  		public int nextInt() throws IOException {
  			return Integer.parseInt(next());
  		}
  	 
  		public double nextDouble() throws IOException {
  			return Double.parseDouble(next());
  		}
  	 
  		public char nextChar() throws IOException {
  			return next().charAt(0);
  		}
  	 
  		public long nextLong() throws IOException {
  			return Long.parseLong(next());
  		}
  	 
  		public boolean ready() throws IOException {
  			return br.ready();
  		}
  	 
  		public void waitForInput() throws InterruptedException {
  			Thread.sleep(3000);
  		}
  		
  	}
  	static void dbg(int[]in) {
  		System.out.println(Arrays.toString(in));
  	}
  	static void dbg(long[]in) {
  		System.out.println(Arrays.toString(in));
  	}
  	static void sort(int[]in) {
  		shuffle(in);
  		Arrays.sort(in);
  	}
  	static void sort(long[]in) {
  		shuffle(in);
  		Arrays.sort(in);
  	}
  	static void shuffle(int[]in) {
  		for(int i=0;i<in.length;i++) {
  			int idx=(int)(Math.random()*in.length);
  			int tmp=in[i];
  			in[i]=in[idx];
  			in[idx]=tmp;
  		}
  	}
  	static void shuffle(long[]in) {
  		for(int i=0;i<in.length;i++) {
  			int idx=(int)(Math.random()*in.length);
  			long tmp=in[i];
  			in[i]=in[idx];
  			in[idx]=tmp;
  		}
  	}
  }