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.math.*;
  import java.security.*;
  import java.text.*;
  import java.time.LocalTime;
  import java.util.*;
  import java.util.concurrent.*;
  import java.util.regex.*;
  
  
  public class A {
  	
  
  	public static void main(String[] args) {
      	FastScanner scan = new FastScanner();
      	//PrintWriter out = new PrintWriter(System.out);
      	
      	int T = 1;
      	T = scan.nextInt();
      		for(int tt=0; tt<T; tt++) {
      			int n = scan.nextInt(), k = scan.nextInt();
      			 a = scan.readArray(n);
      			int [] freq = new int[n+1];
      			ArrayList<Integer> list = new ArrayList<Integer>();
      			for(int i = 0; i<n; i++) {
      				if(freq[a[i]] < k) {
      					freq[a[i]]++;
      					list.add(i);
      				}
      			}
      			int [] ans = new int[n];
      			while(list.size() % k != 0) {
      				list.remove(list.size()-1);
      			}
      			
      			Collections.sort(list, new IntComparator());
      			int currColor = 1;
      			for(int i: list) {
      				ans[i] = currColor++;
      				if(currColor > k) currColor =1 ;
      			}
      			printIntArray(ans);
      		}
  	}
  	
  	static int [] a;
  	
  	static class IntComparator implements Comparator<Integer> {
  
  		public int compare(Integer o1, Integer o2) {
  			return Integer.compare(a[o1], a[o2]);
  		}
  		
  	}
  	
  	
  	public static void printIntArray(int [] a) {
  		for(int i: a) System.out.print(i + " ");
  		System.out.println();
  	}
  	
  	public static void swapInt(int a, int b) {
  		int temp = a;
  		a = b;
  		b = temp;
  	}
  
  	
  	  public static void sort(int [] a) {
  	    	ArrayList<Integer> b = new ArrayList<>();
  	    	for(int i: a) b.add(i);
  	    	Collections.sort(b);
  	    	for(int i=0; i<a.length; i++) a[i]= b.get(i);
  	    }
  	  
  	  public static void reverse(int [] a) {
  	    	ArrayList<Integer> b = new ArrayList<>();
  	    	for(int i: a) b.add(i);
  	    	Collections.reverse(b);
  	    	for(int i=0; i<a.length; i++) a[i]= b.get(i);
  	    }
  
  
  		
  	
  	
      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());
      	}
      	
      	
      }
      
      
     
  }
  
  