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.IOException;
  import java.io.InputStreamReader;
  import java.util.*;
  public class B2_734{
  	public static void main(String[] args){
  		FastReader sc = new FastReader();
  		int t=sc.nextInt();
  		while(t-->0){
  			int n=sc.nextInt();int k=sc.nextInt();
  			int a[]=sc.fastArray(n);
  			HashMap <Integer,ArrayList<Integer>> map = new HashMap<>();
  			for(int i=0;i<n;i++) {
  				if(map.containsKey(a[i])) {
  					map.get(a[i]).add(i);
  				}else {
  					map.put(a[i], new ArrayList<Integer>());
  					map.get(a[i]).add(i);
  				}
  			}
  			int sum=0;
  			for(int i:map.keySet()) {
  				int temp=map.get(i).size();
  				int min=Math.min(k, temp);
  				sum+=min;
  			}
  			int ans[]=new int [n];int colors=0;
  			a:for(int i:map.keySet()) {
  				int temp=map.get(i).size();
  				int min=Math.min(k, temp);
  				for(int j=0;j<min;j++) {
  					colors++;
  					if(colors>k)colors=1;
  					ans[map.get(i).get(j)]=colors;
  					sum--;
  					if(colors==k)
  						if(sum<k) {
  							break a;
  						}
  				}
  			}
  			print(ans);
  			
  		}		 
      }
  
  	static class pair {
  		int x;int y;
  		 pair(int x,int y){
  			this.x=x;
  			this.y=y;
  		}
  		
  	}
  	static void print(int a[]) {
  		for(int i:a) {
  			System.out.print(i+" ");
  		}
  		System.out.println();
  	}
  	static class FastReader {
  		BufferedReader br;
  		StringTokenizer st;
  
  		public FastReader()
  		{
  			br = new BufferedReader(
  				new InputStreamReader(System.in));
  		}
  		int [] fastArray(int n) {
  			int a[]=new int [n];
  			for(int i=0;i<n;i++) {
  				a[i]=nextInt();
  			}
  			return a;
  		}
  		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;
  		}
  	}
  
  }