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 Aqueous {
  		
  		
  		static MyScanner sc = new MyScanner();
  			
  		public static void main(String[] args) {
  			
  			int t = sc.nextInt();
  			
  				while(t-->0) {
  					
  					int n = sc.nextInt();
  					int k = sc.nextInt();
  					
  					int a[] = new int[n];
  					HashMap<Integer, ArrayList<Integer>> hm = new HashMap<Integer, ArrayList<Integer>>();
  					
  						for(int i = 0; i<n; i++) {
  							a[i] = sc.nextInt();
  							
  							if(hm.containsKey(a[i])) {
  								
  								hm.get(a[i]).add(i);
  							}
  							else {
  								hm.put(a[i], new ArrayList<Integer>());
  								hm.get(a[i]).add(i);
  							}
  						}
  						
  						int res[] = new int[n];
  						
  						int col = 1;
  						
  						HashMap<Integer, ArrayList<Integer>> hh = new HashMap<Integer, ArrayList<Integer>>();
  						
  						for(int i =1; i<=k; i++) {
  							hh.put(i, new ArrayList<Integer>());
  						}
  						
  						for(int e: hm.keySet()) {
  							
  							int cnt = 0;
  							
  							for(int i = 0; i<hm.get(e).size(); i++) {
  								
  								if(cnt>=k) {
  									break;
  								}
  								
  								res[hm.get(e).get(i)] = col;
  								
  								hh.get(col).add(hm.get(e).get(i));
  								
  								col++;
  								cnt++;
  								
  								if(col>k) {
  									col=1;
  								}
  							}
  						}
  						
  						int mn = Integer.MAX_VALUE;
  						
  						for(int i =1; i<=k; i++) {
  							if(hh.get(i).size()<mn) {
  								mn = hh.get(i).size();
  							}
  						}
  						
  						for(int x : hh.keySet()) {
  							
  							int r = hh.get(x).size()-mn;
  								
  								for(int i = 0; i<r; i++) {
  									res[hh.get(x).get(i)]=0;
  								}
  						}
  						
  						for(int e: res) {
  							System.out.print(e+" ");
  						}
  						System.out.println();
  						
  				}
  		}
  		
  		static void ruffleSort(int a[]) {
  
  	          int n = a.length;
  
  	          Random r = new Random();
  
  	          for (int i = 0; i < n; i++) {
  
  	              int oi = r.nextInt(n);
  	              int temp = a[oi];
  	              a[oi] = a[i];
  	              a[i] = temp;
  	          }
  
  	          Arrays.sort(a);
  	      }
  		
  		
  		static void ruffleSort(long a[]) {
  
  	          int n = a.length;
  
  	          Random r = new Random();
  
  	          for (int i = 0; i < n; i++) {
  	              int oi = r.nextInt(n);
  
  	              long temp = a[oi];
  	              a[oi] = a[i];
  	              a[i] = temp;
  	          }
  
  	          Arrays.sort(a);
  	      }
  
  		
  		
  	    public static class MyScanner {
  	        BufferedReader br;
  	        StringTokenizer st;
  
  	        public MyScanner() {
  	            br = new BufferedReader(new InputStreamReader(System.in));
  	        }
  
  	        String next() {
  	            while (st == null || !st.hasMoreElements()) {
  	                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());
  	        }
  
  	        String nextLine() {
  	            String str = "";
  	            try {
  	                str = br.readLine();
  	            } catch (IOException e) {
  	                e.printStackTrace();
  	            }
  	            return str;
  	        }
  
  	    }
  	}
  