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: /*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
  
  */
  import java.util.*;
  import java.lang.*;
  import java.io.*;
  
  public class Main
  {
  	static PrintWriter out;
  	static int MOD = 1000000007;
  	static FastReader scan;
  
      /*-------- I/O usaing short named function ---------*/
  	public static String ns(){return scan.next();}
  	public static int ni(){return scan.nextInt();}
  	public static long nl(){return scan.nextLong();}
  	public static double nd(){return scan.nextDouble();}
  	public static String nln(){return scan.nextLine();}
  	public static void p(Object o){out.print(o);}
  	public static void ps(Object o){out.print(o + " ");}
  	public static void pn(Object o){out.println(o);}
      /*-------- for output of an array ---------------------*/
       static void iPA(int arr []){
       	 StringBuilder output = new StringBuilder();
  	     for(int i=0; i<arr.length; i++)output.append(arr[i] + " ");out.println(output);
  	}
  	static void lPA(long arr []){
  	     StringBuilder output = new StringBuilder();
  	     for(int i=0; i<arr.length; i++)output.append(arr[i] + " ");out.println(output);
  	}
  	static void sPA(String arr []){
  	     StringBuilder output = new StringBuilder();
  	     for(int i=0; i<arr.length; i++)output.append(arr[i] + " ");out.println(output);
  	}
  	static void dPA(double arr []){
  	     StringBuilder output = new StringBuilder();
  	     for(int i=0; i<arr.length; i++)output.append(arr[i] + " ");out.println(output);
  	}
      /*-------------- for input in an array ---------------------*/
  	static void iIA(int arr[]){
  	     for(int i=0; i<arr.length; i++)arr[i] = ni();
  	}
  	static void lIA(long arr[]){
  	     for(int i=0; i<arr.length; i++)arr[i] = nl(); 
  	}
  	static void sIA(String arr[]){
  	     for(int i=0; i<arr.length; i++)arr[i] = ns(); 
  	}
  	static void dIA(double arr[]){
  	     for(int i=0; i<arr.length; i++)arr[i] = nd();
  	}
  	/*------------ for taking input faster ----------------*/
       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;
  		}
  	}
  
  	// Method to check if x is power of 2
      static boolean isPowerOfTwo (int x) { return x!=0 && ((x&(x-1)) == 0);}
      //Method to return gcd of two numbers
      static int gcd(int a, int b){return a==0?b:gcd(b % a, a); }
      //Method to count digit of a number
      static int countDigit(long n){return (int)Math.floor(Math.log10(n) + 1);}
      //Method for sorting
  	static void sort(int[] a) {
  		ArrayList<Integer> l=new ArrayList<>();
  		for (int i:a) l.add(i);
  		Collections.sort(l);
  		for (int i=0; i<a.length; i++) a[i]=l.get(i);
  	}
  	//Method for sorting
  	static void sort(long[] a) {
  		ArrayList<Long> l=new ArrayList<>();
  		for (long i:a) l.add(i);
  		Collections.sort(l);
  		for (int i=0; i<a.length; i++) a[i]=l.get(i);
  	}
  	//Method foAr checking if a number is prime or not
  	static boolean isPrime(int n) { 
          if (n <= 1) return false; 
          if (n <= 3) return true; 
          if (n % 2 == 0 || n % 3 == 0) return false; 
          for (int i = 5; i * i <= n; i = i + 6) 
              if (n % i == 0 || n % (i + 2) == 0) 
              	return false; 
          return true; 
      } 
  
  	public static void main (String[] args) throws java.lang.Exception
  	{
          out =new PrintWriter(System.out);
          scan =new FastReader();
  		//for fast output sometimes
  		StringBuilder sb = new StringBuilder();
  		int t = ni();
  		while(t-->0){
  			int n = ni(), k = ni();
  			int arr[] = new int[n];
  			iIA(arr);
  			int cnt[] = new int[n+1];
  			/*for(int i=0; i<n; i++){
  				cnt[arr[i]]++;
  			}
  			
  			for(int i=0; i<=n; i++){
  				if(cnt[i] >= k){
  					chance = Math.min(chance, k);
  				}
  			}*/
  
  			ArrayList<Integer> adj[] = new ArrayList[n+1];
  			for(int i=1; i<=n; i++){
  				adj[i] = new ArrayList<Integer>();
  			}
  
  			for(int i=0; i<n; i++){
  				adj[arr[i]].add(i);
  			}
  
  			int color[] = new int[n];
  			
  			int group = 0, x= 0;
  			ArrayList<Integer> list = new ArrayList<>();
  
  			for(int i=1; i<=n; i++){
  				group = 0;
  				for(int idx : adj[i]){
  				
  					color[idx] = x + 1;
  					list.add(idx);
  
  					x++;
  					x%= k;
  					group++;
  
  					if(x == 0){
  						list.clear();
  					}
  					if(group>=k){
  						group = 0;
  						break;
  					}
  				}
  			}
  			//pn(list.size());
  			if(list.size() != 0){
  				for(int idx : list){
  					color[idx] = 0;
  				}
  			}
  			iPA(color);
          }
          out.flush();
  		out.close();
  	}
  	
  }