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.*;    //PrintWriter
  import java.math.*;  //BigInteger, BigDecimal
  import java.util.*;  //StringTokenizer, ArrayList
  
  
  public class R734_Div3_C
  {	
  	FastReader in;
  	PrintWriter out;
  	
  	public static void main(String[] args)  {
  		new R734_Div3_C().run();
  	}
  	
  	void run()
  	{		
  		in = new FastReader(System.in);
  		out = new PrintWriter(System.out);
  		solve();
  		out.close();
  	}
  	
  	void solve()
  	{
  		int t = in.nextInt();
  		for (int T = 0; T < t; T++) {
  			int n = in.nextInt();
  			int k = in.nextInt();
  			
  			int[] a = new int[n];
  			int[] c = new int[n];
  			HashMap<Integer,Integer> hm = new HashMap<>();
  			
  			int cnt = 0;
  			ArrayList<Integer> u = new ArrayList<>();
  			
  			ArrayList<Integer>[] al = new ArrayList[n+1]; 
  			for (int i=0;i<=n;i++) al[i] = new ArrayList<Integer>();
  			
  			for (int i = 0; i < n; i++) {				
  				a[i] = in.nextInt();
  				if (hm.containsKey(a[i])) {
  					cnt = hm.get(a[i]) + 1;	
  				} else {
  					cnt = 1;
  					u.add(a[i]);
  				}
  				hm.put(a[i], cnt);
  				al[a[i]].add(i);
  			}
  			
  			int tot = 0, empty = 0;
  			int color = 1, placed = 0;
  			for (int x : u) {
  				int num = hm.get(x);
  				if (num > k) {
  					empty += (num - k);
  				}
  			}
  			tot = n - empty;
  			tot = (tot / k) * k;
  			//out.println(tot);
  			
  			for (int x : u) {
  				int num = hm.get(x);
  				for (int i = 0; i < num && i < k && placed < tot; i++) {
  					c[al[x].get(i)] = color;
  					placed++;
  					color++;
  					if (color > k) color = 1;
  				}
  			}
  			
  			for (int i = 0; i < n; i++) {
  				out.print(c[i] + " ");					
  			}
  			out.println();
  		}
  	}
  
  	//-----------------------------------------------------
  	void runWithFiles() {
  		in = new FastReader(new File("input.txt"));
  		try {
  			out = new PrintWriter(new File("output.txt"));
  		} 
  		catch (FileNotFoundException e) {
  			e.printStackTrace();
  		}
  		solve();
  		out.close();
  	}
  	
  	class FastReader
  	{
  	    BufferedReader br;
  	    StringTokenizer tokenizer;
  	    
  	    public FastReader(InputStream stream)
  	    {
  	        br = new BufferedReader(new InputStreamReader(stream));
  	        tokenizer = null;
  	    }
  		public FastReader(File f) {
  			try {
  				br = new BufferedReader(new FileReader(f));
  				tokenizer = null;
  			}
  			catch (FileNotFoundException e) {
  				e.printStackTrace();
  			}
  		}
  	    
  	    private String next() {
  	        while (tokenizer == null || !tokenizer.hasMoreTokens())
  	            try {
  	            	tokenizer = new StringTokenizer(br.readLine());
  	            }
  	            catch (IOException e) {
  	                throw new RuntimeException(e);
  	            }
  	        return tokenizer.nextToken();
  	    }
  		public String nextLine() {
  			try	{
  				return br.readLine();
  			}
  			catch(Exception e) {
  				throw(new RuntimeException());
  			}
  		}
  
  	    int nextInt() {
  	        return Integer.parseInt(next());
  	    }
  	    long nextLong() {
  	        return Long.parseLong(next());
  	    }
  	    double nextDouble() {
  	        return Double.parseDouble(next());
  	    }	    
  	    BigInteger nextBigInteger() {
  	        return new BigInteger(next());
  	    }
  	    BigDecimal nextBigDecimal() {
  	        return new BigDecimal(next());
  	    }
  	}
  }
  