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 static java.lang.Math.max;
  import static java.lang.Math.min;
  import static java.lang.Math.abs;
  import static java.lang.System.out;
  import java.util.* ;
  import java.math.*;
  import java.io.*;
  
  
   public class javaTemplate {
  	
  	public static final int M = 1000000007 ;
  	static FastReader sc = new FastReader();
  	static int N = 86028121 ;
  	static boolean globB[] = new boolean[N+1] ;
  	static int globA[] = new int[10000000] ;
  	
  	
  	public static void main(String[] args) {
  		
  		int q = sc.nextInt() ;
  		
  		while(q -- != 0) {
  			int n = sc.nextInt() ;
  			int k = sc.nextInt() ;
  			Map<Integer,ArrayList<Integer>> map = new HashMap<>() ;
  			ArrayList<Integer> list = new ArrayList<>() ;
  			int arr[] = new int[n] ;
  			int ans[] = new int[n] ;
  			Arrays.fill(ans, 0);
  			for(int i = 0 ; i< n ; i++) {
  				arr[i] = sc.nextInt() ;
  			}
  			
  //			ArrayList<Integer> li ;
  			for(int i = 0 ; i< n ; i++) {
  				if(map.containsKey(arr[i])) {
  					ArrayList<Integer> li = map.get(arr[i]) ;
  					li.add(i) ;
  					map.put(arr[i],li) ;
  				} else {
  					ArrayList<Integer> li = new ArrayList<>() ;
  					li.add(i) ;
  					map.put(arr[i],li) ;
  				}
  			}
  			
  			int extra = 0 ;
  			for(Map.Entry<Integer, ArrayList<Integer>> e: map.entrySet()) {
  				ArrayList<Integer> x = e.getValue() ;
  				int y = e.getKey() ;
  				int z = 1 ;
  				if(x.size() >= k) {
  					for(int i = 0 ; i< k ; i++) {
  						ans[x.get(i)] = z ;
  						z ++ ;
  					}
  				} else {
  					for(int i = 0 ; i< x.size() ; i++) {
  						list.add(x.get(i)) ;
  					}
  					extra += x.size() ;
  				}
  			}
  			
  			int limit = extra - (extra%k) ;
  			int z = 1 ;
  			for(int i = 0 ; i< limit ; i++) {
  				if(z > k) z = 1 ;
  //				System.out.print(list.get(i) + " ");
  				ans[list.get(i)] = z ;
  				z ++ ;
  			}
  //			System.out.println();
  			for(int e: ans) {
  				System.out.print(e + " ");
  			}
  			
  			
  			System.out.println();
  			
  		}
  
  	}
  
  
  	/*
  	 @ Observation is imp. for problem A and B
  	 1->> Read the problem properly take time to read the problem 
  	 if it is too lengthy.
  	 2->> Draw the test cases of A and B 
  	 3->> Don't over think the problem 
  	 4->> Try to take sum variable as long
  	 5->> Avoid Nested if else statement
  	 6->> If it looks like a Greedy problem try to figure out the ans. 
  	 for the smallest test case.
  	 7->> If it is a binary String check no. of 0's and 1's. 
  	 If it is a even try to pair it (may be not applicable for every questions).
  	 8->> Always check the constrains and try BruteForce
  	 @ Implement it quickly 
  	 */
  	
  
  	
  	
  			
  	
  	
  	
  	
  
  
  	
  	
  //_________________________//Template//_____________________________________________________________________
  	
  	
  		//		FAST I/O
  		static class FastReader {
  	        BufferedReader br;
  	        StringTokenizer st;
  	        
  	        public FastReader()
  	        {
  	            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;
  	        }
  		}
  	
  	
  		
  		
  		
  		// Sorting
  		public static void sort(int[] arr) {
  	        //because Arrays.sort() uses quicksort which is dumb
  	        //Collections.sort() uses merge sort
  	        ArrayList<Integer> ls = new ArrayList<Integer>();
  	        for(int x: arr)
  	            ls.add(x);
  	        Collections.sort(ls);
  	        for(int i=0; i < arr.length; i++)
  	            arr[i] = ls.get(i);
  	    }
  		
  		
  		
  		
  		
  		// Modulo
  		static int mod(int a) {
  		   return (a%M + M) % M;
  		}
  		
  		
  		
  		
  		
  		
  		
  		//		Palindrom or Not
  		static boolean isPalindrome(String str, int low, int high) {
  					while (low < high) {
  						if (str.charAt(low) != str.charAt(high)) return false;
  					 	low++;
  					 	high--;
  					}
  					return true;
  				}
  	
  	
  
  	
  	
  	
  	
  	
  	
  	
  	
  		// check if sorted or not	
  		public static boolean isSorted(int[] a)
  	    {
  	        for (int i = 0; i < a.length - 1; i++)
  	        {
  	            if (a[i] > a[i + 1]) {
  	                return false;
  	            }
  	        }
  	        return true;
  	    }
  	
  	
  	
  		
  		
  	
  		// GCD
  		public static int GCD(int a , int b) {
  				if(b == 0) return a ;
  				
  				return GCD(b, a%b) ;
  			}
  	
  		
  		
  		
  		
  		// sieve
  		public static void sieve() {
  			Arrays.fill(globB, true);
  			globB[0] = false ;
  			globB[1] = false ;
  			for(int i = 2 ; i * i <= N ; i++) {
  				if(globB[i]) {
  				for(int j = i * i ; j<= N ; j+= i) {
  					globB[j] = false ;
  				 }
  				}
  			}
  			
  		}
  
  		
  		
  		
  		// fastPower
  		public static long fastPowerModulo(long a, long b, long n) {
  			long res = 1 ;
  			while(b > 0) {
  			if((b&1) != 0) {
  				res = (res * a % n) % n ;
  			}
  			a = (a % n * a % n) % n ;
  			b = b >> 1 ;
  			}
  			return res ;
  		}
  	
  		
  		
          // LCM
  		static int LCM(int a, int b)
  	    {
  	        return (a / GCD(a, b)) * b;
  	    }
  		
  
  }
  
  
   
   
   
   
   
   
  