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.util.*;
  
  public class WonderfulColoring2 {
  
  	public static void main(String[] args) throws IOException {
  		// TODO Auto-generated method stub
  		// System.in and System.out are input and output streams, respectively.
  		InputStream inputStream = System.in;
  
  		InputReader in = new InputReader(inputStream);
  		BufferedWriter out = new BufferedWriter(new OutputStreamWriter(System.out));
  		int t = in.nextInt();
  		while (t-- != 0) {
  			int n = in.nextInt();
  			int k = in.nextInt();
  			HashMap<Integer, Queue<Integer>> pos = new HashMap<>();
  			int total = 0;
  			for (int idx = 0; idx < n; idx++) {
  				int num = in.nextInt();
  				if (!pos.containsKey(num)) {
  					pos.put(num, new LinkedList<>());
  				}
  				if (pos.get(num).size() < k) {
  					pos.get(num).offer(idx);
  					total++;
  				}
  			}
  			Set<Integer> set = pos.keySet();
  			int[] ans = new int[n];
  			int color = 1;
  			boolean shouldBreak = false;
  			for (int x : set) {
  				Queue<Integer> queue = pos.get(x);
  				while (!queue.isEmpty()) {
  					ans[queue.poll()] = color;
  					color++;
  					total--;
  					if (color == k + 1) {
  						color = 1;
  						if (total - k < 0) {
  							shouldBreak = true;
  							break;
  						}
  					}
  				}
  				if(shouldBreak) break;
  			}
  
  			for (int x : ans) {
  				out.write(x + " ");
  			}
  			out.newLine();
  		}
  		out.close();
  	}
  
  	static class InputReader {
  		public BufferedReader reader;
  		public StringTokenizer tokenizer;
  
  		public InputReader(InputStream stream) {
  			reader = new BufferedReader(new InputStreamReader(stream), 32768);
  			tokenizer = null;
  		}
  
  		public String next() {
  			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
  				try {
  					tokenizer = new StringTokenizer(reader.readLine());
  				} catch (IOException e) {
  					throw new RuntimeException(e);
  				}
  			}
  			return tokenizer.nextToken();
  		}
  
  		public int nextInt() {
  			return Integer.parseInt(next());
  		}
  
  		public long nextLong() {
  			return Long.parseLong(next());
  		}
  
  		public double nextDouble() {
  			return Double.parseDouble(next());
  		}
  
  		public int[] readArray(int n) {
  			int[] arr = new int[n];
  			for (int i = 0; i < n; i++)
  				arr[i] = nextInt();
  			return arr;
  		}
  	}
  }
  