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.*;
  import java.io.*;
  public class B{
  	public static void main(String[] args)
  	{
  		FastScanner fs = new FastScanner();
  		PrintWriter out = new PrintWriter(System.out);
  		int t = fs.nextInt();
  		for(int tt=0;tt<t;tt++)
  		{
  			int n = fs.nextInt(); int k = fs.nextInt();
  			
  			int[] arr = fs.readArray(n);
  			
  			int max = 0;
  			for(int i:arr)max = Math.max(max,i);
  			
  			List<Integer>[]adj = new ArrayList[max+1];
  			for(int i=0;i<=max;i++)adj[i] = new ArrayList();
  			int[] vis = new int[max+1];
  			for(int i=0;i<n;i++)
  			{
  				adj[arr[i]].add(i);
  			}
  			
  			int[] ans = new int[n];
  			
  			int size = 0;
  			
  			for(int i=0;i<n;i++)
  			{
  				if(vis[arr[i]] == 0) {
  				if(adj[arr[i]].size()>k)
  				{
  					size += k;
  				}
  				else size += adj[arr[i]].size();
  				vis[arr[i]] = 1;
  //				size += (Math.min(k,adj[arr[i]].size()));
  				}
  			}
  //			System.out.println(size);
  			size/=k;
  			
  			adj = new ArrayList[max+1];
  			
  			for(int i=0;i<=max;i++)adj[i] = new ArrayList();
  			
  			for(int i=0;i<n;i++)
  			{
  				adj[arr[i]].add(i);
  			}
  //			System.out.println(size);
  			int start = 1;
  			int done = 0;
  			me:for(int i=0;i<=max;i++)
  			{
  				if(done>=size)break;
  				int can = Math.min(adj[i].size(),k);
  				for(int ele:adj[i])
  				{
  					if(can<=0)break;
  					if(start == k)
  					{
  //						System.out.println("Here");
  						done++;
  						ans[ele] = start;
  						if(done>=size)break me;
  						start = 1;
  					}
  					else
  					{
  //						System.out.println("here");
  						ans[ele] = start;
  						start++;
  					}
  					can--;
  				}
  			}
  			
  			for(int i:ans)
  			{
  				out.print(i+" ");
  			}
  			out.println();
  			
  		}
  		out.close();
  	}
  	static class FastScanner {
  		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
  		StringTokenizer st=new StringTokenizer("");
  		String next() {
  			while (!st.hasMoreTokens())
  				try {
  					st=new StringTokenizer(br.readLine());
  				} catch (IOException e) {
  					e.printStackTrace();
  				}
  			return st.nextToken();
  		}
  		
  		int nextInt() {
  			return Integer.parseInt(next());
  		}
  		int[] readArray(int n) {
  			int[] a=new int[n];
  			for (int i=0; i<n; i++) a[i]=nextInt();
  			return a;
  		}
  		long nextLong() {
  			return Long.parseLong(next());
  		}
  	}
  	public static int[] sort(int[] arr)
  	{
  		List<Integer> temp = new ArrayList();
  		for(int i:arr)temp.add(i);
  		Collections.sort(temp);
  		int start = 0;
  		for(int i:temp)arr[start++]=i;
  		return arr;
  	}
  	public static String rev(String str)
  	{
  		char[] arr = str.toCharArray();
  		char[] ret = new char[arr.length];
  		int start = arr.length-1;
  		for(char i:arr)ret[start--] = i;
  		String ret1 = "";
  		for(char ch:ret)ret1+=ch;
  		return ret1;
  	}
  }