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.PrintWriter;
  import java.io.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.util.*;
  public class solution { 
      public static void main(String args[]) throws java.lang.Exception{ 
      	FastScanner s=new FastScanner();
  		PrintWriter out=new PrintWriter(System.out);
  		int t=s.nextInt();
  		for(int tt=0;tt<t;tt++) {
  			int n=s.nextInt(),k=s.nextInt();
  			int a[]=s.readArray(n);
  			int ans[]=new int[n];
  			HashMap<Integer,ArrayList<Integer>> map=new HashMap<>();
  			for(int i=0;i<n;i++) {
  				if(map.containsKey(a[i])) {
  					ArrayList<Integer> d=map.get(a[i]);
  					d.add(i);
  					map.put(a[i],d);
  				}else {
  					ArrayList<Integer> d=new ArrayList<>();
  					d.add(i);
  					map.put(a[i],d);
  				}
  			}
  			int c1=1;
  			for (ArrayList<Integer> url : map.values()) {
  				ArrayList<Integer> temp=url;
  				for(int i=0;i<Math.min(k,temp.size());i++) {
  					ans[temp.get(i)]=c1;
  					c1++;
  					if(c1==k+1) {
  						c1=1;
  					}
  				}
  			}
  			if(c1!=1) {
  				HashSet<Integer> set=new HashSet<Integer>();
  				for(int i=1;i<c1;i++) {
  					set.add(i);
  				}
  				for(int i=0;i<n;i++) {
  					if(set.contains(ans[i])) {
  						set.remove(ans[i]);
  						ans[i]=0;
  					}
  				}
  			}
  			for(int i=0;i<n;i++) {
  				out.print(ans[i]+" ");
  			}
  			out.println();
  		}
  		out.close();
      }
      public static int modulus(int a,int b) {
      	int ans=a%b;
      	if(ans<0) {
      		ans=b+ans;
      	}
      	return ans;
      }
      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);
  	}
      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);
  	}
      static int gcd(int a, int b){
        if (b == 0)
          return a;
        return gcd(b, a % b);
      }
      static void sortcol(int a[][],int c) {
  		Arrays.sort(a, (x, y) -> {
  			if(x[c]>y[c]) {
  				return 1;
  			}else {
  				return -1;
  			}
  		});
  	}
      public static void printb(boolean ans) {
      	if(ans) {
      		System.out.println("Yes");
      	}else {
      		System.out.println("No");
      	}
      }
      static 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();
  		}
  		double nextDouble() {
  			return Double.parseDouble(next());
  		}
  		int nextInt() {
  			return Integer.parseInt(next());
  		}
  		int[] readArray(int n) {
  			int[] a=new int[n];
  			for (int i=0; i<n; i++) a[i]=nextInt();
  			return a;
  		}
  		long nextLong() {
  			return Long.parseLong(next());
  		}
  	}
      static class Pair implements Comparable<Pair>{
      	int a , b; 
      	Pair(int x , int y){
      		a=x;
      		b=y;
      	}
      	public int compareTo(Pair o) {
      		return a != o.a ? a - o.a : b - o.b;
      	}
      }
  }