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.IOException;
  import java.io.BufferedReader;
  import java.io.InputStreamReader;
  import java.util.StringTokenizer;
  import java.util.HashMap;
  import java.util.ArrayList;
  
  public class a {
  	final static FastScanner scan = new FastScanner();
  	
  	static void test() {
  		int n = scan.nextInt();
  		int k = scan.nextInt();
  		HashMap<Integer, ArrayList<Integer>> map = new HashMap<>();
  		for(int i = 0; i < n; i ++) {
  			int a = scan.nextInt();
  			if(map.containsKey(a)) {
  				;
  			}
  			else {
  				map.put(a, new ArrayList<Integer>());
  			}
  			map.get(a).add(i);
  		}
  		int answer[] = new int[n];
  		int color = 1;
  		int count = 0;
  		for(Integer element : map.keySet()) {
  			count += Math.min(k, map.get(element).size());
  		}
  		count = count / k * k;
  		outer :
  			for(Integer element : map.keySet()) {
  				int size = Math.min(map.get(element).size(), k);
  				for(int i = 0; i < size; i ++) {
  					answer[map.get(element).get(i)] = color;
  					color = color % k + 1;
  					count --;
  					if(count == 0) break outer;
  				}
  			}
  
  		for(int i = 0; i < n; i ++) {
  			System.out.print(answer[i] + " ");
  		}
  		System.out.println();
  	}
  	
  	public static void main(String[] args) {
  		int T = 1;
  		T = scan.nextInt();
  		for(int tc = 0; tc < T; ++ tc) {
  			test();
  		}	
  	}
  }
  
  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());
  	}
  	long nextLong() {
  		return Long.parseLong(next());
  	}
  	double nextDouble() {
  		return Double.parseDouble(next());
  	}
  }
  